Inequality $sumlimits_{cyc}frac{a^3}{13a^2+5b^2}geqfrac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5c^2}+frac{c^3}{13c^2+5a^2}geqfrac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{2a^2+b^2}+frac{b^3}{2b^2+c^2}+frac{c^3}{2c^2+a^2}geqfrac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
real-analysis inequality contest-math
|
show 21 more comments
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5c^2}+frac{c^3}{13c^2+5a^2}geqfrac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{2a^2+b^2}+frac{b^3}{2b^2+c^2}+frac{c^3}{2c^2+a^2}geqfrac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
real-analysis inequality contest-math
2
Is the function $f : (a,b,c) rightarrow sumlimits_{cyc} frac{6a^3}{13a^2+5b^2} $ concave? Also, C-S = "Cauchy-Schwarz inequality"? Avoid this kind of unconventional abbreviation, please.
– Vincent
May 8 '16 at 18:17
See also math.stackexchange.com/questions/1775572/…
– David Quinn
May 8 '16 at 18:42
1
@Santropedro There is the following known inequality $sumlimits_{cyc}frac{a^3}{a^2+b^2}geqfrac{a+b+c}{2}$. I think, it's just interesting, for which maximal value of $k$ the following inequality is still true. $sumlimits_{cyc}frac{a^3}{ka^2+b^2}geqfrac{a+b+c}{k+1}$.
– Michael Rozenberg
Apr 19 '17 at 17:28
1
@Santropedro The value $,k=2.6,$ in the OP is pretty sharp. I found that $k=2.603279$ is already beyond: Then Michael's given $(0.785, 1.25, 1.861)$ violates the ineq..
– Hanno
May 11 '17 at 18:24
1
Further result obtained via random-based simulation: $(0.785, 1.2535, 1.873)$ doesn't satisfy the inequality if $,k=2.603262,$. Below that $k$-value I didn't detect a violation.
– Hanno
May 11 '17 at 18:50
|
show 21 more comments
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5c^2}+frac{c^3}{13c^2+5a^2}geqfrac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{2a^2+b^2}+frac{b^3}{2b^2+c^2}+frac{c^3}{2c^2+a^2}geqfrac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
real-analysis inequality contest-math
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5c^2}+frac{c^3}{13c^2+5a^2}geqfrac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{2a^2+b^2}+frac{b^3}{2b^2+c^2}+frac{c^3}{2c^2+a^2}geqfrac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
real-analysis inequality contest-math
real-analysis inequality contest-math
edited Dec 8 at 12:34
asked May 8 '16 at 18:10
Michael Rozenberg
95.4k1588183
95.4k1588183
2
Is the function $f : (a,b,c) rightarrow sumlimits_{cyc} frac{6a^3}{13a^2+5b^2} $ concave? Also, C-S = "Cauchy-Schwarz inequality"? Avoid this kind of unconventional abbreviation, please.
– Vincent
May 8 '16 at 18:17
See also math.stackexchange.com/questions/1775572/…
– David Quinn
May 8 '16 at 18:42
1
@Santropedro There is the following known inequality $sumlimits_{cyc}frac{a^3}{a^2+b^2}geqfrac{a+b+c}{2}$. I think, it's just interesting, for which maximal value of $k$ the following inequality is still true. $sumlimits_{cyc}frac{a^3}{ka^2+b^2}geqfrac{a+b+c}{k+1}$.
– Michael Rozenberg
Apr 19 '17 at 17:28
1
@Santropedro The value $,k=2.6,$ in the OP is pretty sharp. I found that $k=2.603279$ is already beyond: Then Michael's given $(0.785, 1.25, 1.861)$ violates the ineq..
– Hanno
May 11 '17 at 18:24
1
Further result obtained via random-based simulation: $(0.785, 1.2535, 1.873)$ doesn't satisfy the inequality if $,k=2.603262,$. Below that $k$-value I didn't detect a violation.
– Hanno
May 11 '17 at 18:50
|
show 21 more comments
2
Is the function $f : (a,b,c) rightarrow sumlimits_{cyc} frac{6a^3}{13a^2+5b^2} $ concave? Also, C-S = "Cauchy-Schwarz inequality"? Avoid this kind of unconventional abbreviation, please.
– Vincent
May 8 '16 at 18:17
See also math.stackexchange.com/questions/1775572/…
– David Quinn
May 8 '16 at 18:42
1
@Santropedro There is the following known inequality $sumlimits_{cyc}frac{a^3}{a^2+b^2}geqfrac{a+b+c}{2}$. I think, it's just interesting, for which maximal value of $k$ the following inequality is still true. $sumlimits_{cyc}frac{a^3}{ka^2+b^2}geqfrac{a+b+c}{k+1}$.
– Michael Rozenberg
Apr 19 '17 at 17:28
1
@Santropedro The value $,k=2.6,$ in the OP is pretty sharp. I found that $k=2.603279$ is already beyond: Then Michael's given $(0.785, 1.25, 1.861)$ violates the ineq..
– Hanno
May 11 '17 at 18:24
1
Further result obtained via random-based simulation: $(0.785, 1.2535, 1.873)$ doesn't satisfy the inequality if $,k=2.603262,$. Below that $k$-value I didn't detect a violation.
– Hanno
May 11 '17 at 18:50
2
2
Is the function $f : (a,b,c) rightarrow sumlimits_{cyc} frac{6a^3}{13a^2+5b^2} $ concave? Also, C-S = "Cauchy-Schwarz inequality"? Avoid this kind of unconventional abbreviation, please.
– Vincent
May 8 '16 at 18:17
Is the function $f : (a,b,c) rightarrow sumlimits_{cyc} frac{6a^3}{13a^2+5b^2} $ concave? Also, C-S = "Cauchy-Schwarz inequality"? Avoid this kind of unconventional abbreviation, please.
– Vincent
May 8 '16 at 18:17
See also math.stackexchange.com/questions/1775572/…
– David Quinn
May 8 '16 at 18:42
See also math.stackexchange.com/questions/1775572/…
– David Quinn
May 8 '16 at 18:42
1
1
@Santropedro There is the following known inequality $sumlimits_{cyc}frac{a^3}{a^2+b^2}geqfrac{a+b+c}{2}$. I think, it's just interesting, for which maximal value of $k$ the following inequality is still true. $sumlimits_{cyc}frac{a^3}{ka^2+b^2}geqfrac{a+b+c}{k+1}$.
– Michael Rozenberg
Apr 19 '17 at 17:28
@Santropedro There is the following known inequality $sumlimits_{cyc}frac{a^3}{a^2+b^2}geqfrac{a+b+c}{2}$. I think, it's just interesting, for which maximal value of $k$ the following inequality is still true. $sumlimits_{cyc}frac{a^3}{ka^2+b^2}geqfrac{a+b+c}{k+1}$.
– Michael Rozenberg
Apr 19 '17 at 17:28
1
1
@Santropedro The value $,k=2.6,$ in the OP is pretty sharp. I found that $k=2.603279$ is already beyond: Then Michael's given $(0.785, 1.25, 1.861)$ violates the ineq..
– Hanno
May 11 '17 at 18:24
@Santropedro The value $,k=2.6,$ in the OP is pretty sharp. I found that $k=2.603279$ is already beyond: Then Michael's given $(0.785, 1.25, 1.861)$ violates the ineq..
– Hanno
May 11 '17 at 18:24
1
1
Further result obtained via random-based simulation: $(0.785, 1.2535, 1.873)$ doesn't satisfy the inequality if $,k=2.603262,$. Below that $k$-value I didn't detect a violation.
– Hanno
May 11 '17 at 18:50
Further result obtained via random-based simulation: $(0.785, 1.2535, 1.873)$ doesn't satisfy the inequality if $,k=2.603262,$. Below that $k$-value I didn't detect a violation.
– Hanno
May 11 '17 at 18:50
|
show 21 more comments
2 Answers
2
active
oldest
votes
Your inequality is equivalent to :
$$sum_{cyc}frac{a}{13}sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2geq frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$frac{a}{13b} sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2+frac{1}{13}sin(arctan(sqrt{frac{13}{5}}frac{b}{c}))^2+frac{c}{13b}sin(arctan(sqrt{frac{13}{5}}frac{c}{a}))^2geq frac{1+frac{a}{b}+frac{c}{b}}{18}$$
Now we put $sqrt{frac{13}{5}}frac{a}{b}=x$, $sqrt{frac{13}{5}}frac{b}{c}=y$, $sqrt{frac{13}{5}}frac{c}{a}=z$, your inequality is equivalent to:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}$$$$geq dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
with the condition $xyz=(sqrt{frac{13}{5}})^3$.
We study the following function:
$$f(x)=sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}-dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
This function is easily differentiable and the minimum is for $x=sqrt{frac{3sqrt{77}}{17}-frac{26}{17}}=alpha$. So with the condition $xyz=(sqrt{frac{13}{5}})^3$ becomes $yz=frac{(sqrt{frac{13}{5}})^3}{alpha}=beta$. So we have this inequality just with $y$:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(alpha)^3}{(alpha^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(frac{beta}{y})^2}{((frac{beta}{y})^2+1)}geqdfrac{1+sqrt{dfrac{5}{13}}alpha+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
which is easily analyzable. Done!
Why the last inequality is true?
– Michael Rozenberg
Oct 21 '17 at 14:43
Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think .
– max8128
Oct 21 '17 at 14:45
We can check the starting inequality by WA without your previous work. But maybe WA is wrong (by the way, it happens sometimes!).
– Michael Rozenberg
Oct 21 '17 at 15:02
1
Delete please your "solution". I think it's not good, which you are doing.
– Michael Rozenberg
Oct 21 '17 at 16:59
4
I think it's nothing.
– Michael Rozenberg
Oct 25 '17 at 11:15
|
show 3 more comments
I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5a^2}geq frac{a+b}{18}$$
Proof:
We have with $x=frac{a}{b}$ :
$$frac{x^3}{13x^2+5}+frac{1}{13+5x^2}geq frac{1+x}{18}$$
Or
$$5(x+1)(x-1)^2(5x^2-8x+5)geq 0$$
So we have (if we permute the variables $a,b,c$ and addition the three inequalities ) :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}+sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{9}$$
If we have $sum_{cyc}frac{a^3}{13a^2+5b^2}geqsum_{cyc}frac{a^3}{13a^2+5c^2}$
We have :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$$
But also
$$frac{(a-epsilon)^3}{13(a-epsilon)^2+5b^2}+frac{(b)^3}{13(b)^2+5(c+epsilon)^2}+frac{(c+epsilon)^3}{13(c+epsilon)^2+5(a-epsilon)^2}geq frac{a+b+c}{18}$$
If we put $ageq c $ and $epsilon=a-c$
We finally obtain :
$$sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$$
So all the cases are here so it's proved !
1
If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions.
– Michael Rozenberg
Dec 7 at 15:38
Now it's right and proved .
– max8128
Dec 8 at 11:07
I can only see that you proved that, for a given triple $(a, b, c)$, $sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$ or $sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$ must hold. – The same applies to your other answer math.stackexchange.com/a/3041504/42969.
– Martin R
Dec 15 at 15:49
add a comment |
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2 Answers
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Your inequality is equivalent to :
$$sum_{cyc}frac{a}{13}sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2geq frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$frac{a}{13b} sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2+frac{1}{13}sin(arctan(sqrt{frac{13}{5}}frac{b}{c}))^2+frac{c}{13b}sin(arctan(sqrt{frac{13}{5}}frac{c}{a}))^2geq frac{1+frac{a}{b}+frac{c}{b}}{18}$$
Now we put $sqrt{frac{13}{5}}frac{a}{b}=x$, $sqrt{frac{13}{5}}frac{b}{c}=y$, $sqrt{frac{13}{5}}frac{c}{a}=z$, your inequality is equivalent to:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}$$$$geq dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
with the condition $xyz=(sqrt{frac{13}{5}})^3$.
We study the following function:
$$f(x)=sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}-dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
This function is easily differentiable and the minimum is for $x=sqrt{frac{3sqrt{77}}{17}-frac{26}{17}}=alpha$. So with the condition $xyz=(sqrt{frac{13}{5}})^3$ becomes $yz=frac{(sqrt{frac{13}{5}})^3}{alpha}=beta$. So we have this inequality just with $y$:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(alpha)^3}{(alpha^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(frac{beta}{y})^2}{((frac{beta}{y})^2+1)}geqdfrac{1+sqrt{dfrac{5}{13}}alpha+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
which is easily analyzable. Done!
Why the last inequality is true?
– Michael Rozenberg
Oct 21 '17 at 14:43
Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think .
– max8128
Oct 21 '17 at 14:45
We can check the starting inequality by WA without your previous work. But maybe WA is wrong (by the way, it happens sometimes!).
– Michael Rozenberg
Oct 21 '17 at 15:02
1
Delete please your "solution". I think it's not good, which you are doing.
– Michael Rozenberg
Oct 21 '17 at 16:59
4
I think it's nothing.
– Michael Rozenberg
Oct 25 '17 at 11:15
|
show 3 more comments
Your inequality is equivalent to :
$$sum_{cyc}frac{a}{13}sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2geq frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$frac{a}{13b} sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2+frac{1}{13}sin(arctan(sqrt{frac{13}{5}}frac{b}{c}))^2+frac{c}{13b}sin(arctan(sqrt{frac{13}{5}}frac{c}{a}))^2geq frac{1+frac{a}{b}+frac{c}{b}}{18}$$
Now we put $sqrt{frac{13}{5}}frac{a}{b}=x$, $sqrt{frac{13}{5}}frac{b}{c}=y$, $sqrt{frac{13}{5}}frac{c}{a}=z$, your inequality is equivalent to:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}$$$$geq dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
with the condition $xyz=(sqrt{frac{13}{5}})^3$.
We study the following function:
$$f(x)=sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}-dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
This function is easily differentiable and the minimum is for $x=sqrt{frac{3sqrt{77}}{17}-frac{26}{17}}=alpha$. So with the condition $xyz=(sqrt{frac{13}{5}})^3$ becomes $yz=frac{(sqrt{frac{13}{5}})^3}{alpha}=beta$. So we have this inequality just with $y$:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(alpha)^3}{(alpha^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(frac{beta}{y})^2}{((frac{beta}{y})^2+1)}geqdfrac{1+sqrt{dfrac{5}{13}}alpha+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
which is easily analyzable. Done!
Why the last inequality is true?
– Michael Rozenberg
Oct 21 '17 at 14:43
Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think .
– max8128
Oct 21 '17 at 14:45
We can check the starting inequality by WA without your previous work. But maybe WA is wrong (by the way, it happens sometimes!).
– Michael Rozenberg
Oct 21 '17 at 15:02
1
Delete please your "solution". I think it's not good, which you are doing.
– Michael Rozenberg
Oct 21 '17 at 16:59
4
I think it's nothing.
– Michael Rozenberg
Oct 25 '17 at 11:15
|
show 3 more comments
Your inequality is equivalent to :
$$sum_{cyc}frac{a}{13}sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2geq frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$frac{a}{13b} sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2+frac{1}{13}sin(arctan(sqrt{frac{13}{5}}frac{b}{c}))^2+frac{c}{13b}sin(arctan(sqrt{frac{13}{5}}frac{c}{a}))^2geq frac{1+frac{a}{b}+frac{c}{b}}{18}$$
Now we put $sqrt{frac{13}{5}}frac{a}{b}=x$, $sqrt{frac{13}{5}}frac{b}{c}=y$, $sqrt{frac{13}{5}}frac{c}{a}=z$, your inequality is equivalent to:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}$$$$geq dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
with the condition $xyz=(sqrt{frac{13}{5}})^3$.
We study the following function:
$$f(x)=sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}-dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
This function is easily differentiable and the minimum is for $x=sqrt{frac{3sqrt{77}}{17}-frac{26}{17}}=alpha$. So with the condition $xyz=(sqrt{frac{13}{5}})^3$ becomes $yz=frac{(sqrt{frac{13}{5}})^3}{alpha}=beta$. So we have this inequality just with $y$:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(alpha)^3}{(alpha^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(frac{beta}{y})^2}{((frac{beta}{y})^2+1)}geqdfrac{1+sqrt{dfrac{5}{13}}alpha+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
which is easily analyzable. Done!
Your inequality is equivalent to :
$$sum_{cyc}frac{a}{13}sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2geq frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$frac{a}{13b} sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2+frac{1}{13}sin(arctan(sqrt{frac{13}{5}}frac{b}{c}))^2+frac{c}{13b}sin(arctan(sqrt{frac{13}{5}}frac{c}{a}))^2geq frac{1+frac{a}{b}+frac{c}{b}}{18}$$
Now we put $sqrt{frac{13}{5}}frac{a}{b}=x$, $sqrt{frac{13}{5}}frac{b}{c}=y$, $sqrt{frac{13}{5}}frac{c}{a}=z$, your inequality is equivalent to:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}$$$$geq dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
with the condition $xyz=(sqrt{frac{13}{5}})^3$.
We study the following function:
$$f(x)=sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}-dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
This function is easily differentiable and the minimum is for $x=sqrt{frac{3sqrt{77}}{17}-frac{26}{17}}=alpha$. So with the condition $xyz=(sqrt{frac{13}{5}})^3$ becomes $yz=frac{(sqrt{frac{13}{5}})^3}{alpha}=beta$. So we have this inequality just with $y$:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(alpha)^3}{(alpha^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(frac{beta}{y})^2}{((frac{beta}{y})^2+1)}geqdfrac{1+sqrt{dfrac{5}{13}}alpha+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
which is easily analyzable. Done!
edited Dec 8 at 11:14
Saad
19.7k92252
19.7k92252
answered Oct 21 '17 at 14:41
max8128
1,082421
1,082421
Why the last inequality is true?
– Michael Rozenberg
Oct 21 '17 at 14:43
Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think .
– max8128
Oct 21 '17 at 14:45
We can check the starting inequality by WA without your previous work. But maybe WA is wrong (by the way, it happens sometimes!).
– Michael Rozenberg
Oct 21 '17 at 15:02
1
Delete please your "solution". I think it's not good, which you are doing.
– Michael Rozenberg
Oct 21 '17 at 16:59
4
I think it's nothing.
– Michael Rozenberg
Oct 25 '17 at 11:15
|
show 3 more comments
Why the last inequality is true?
– Michael Rozenberg
Oct 21 '17 at 14:43
Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think .
– max8128
Oct 21 '17 at 14:45
We can check the starting inequality by WA without your previous work. But maybe WA is wrong (by the way, it happens sometimes!).
– Michael Rozenberg
Oct 21 '17 at 15:02
1
Delete please your "solution". I think it's not good, which you are doing.
– Michael Rozenberg
Oct 21 '17 at 16:59
4
I think it's nothing.
– Michael Rozenberg
Oct 25 '17 at 11:15
Why the last inequality is true?
– Michael Rozenberg
Oct 21 '17 at 14:43
Why the last inequality is true?
– Michael Rozenberg
Oct 21 '17 at 14:43
Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think .
– max8128
Oct 21 '17 at 14:45
Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think .
– max8128
Oct 21 '17 at 14:45
We can check the starting inequality by WA without your previous work. But maybe WA is wrong (by the way, it happens sometimes!).
– Michael Rozenberg
Oct 21 '17 at 15:02
We can check the starting inequality by WA without your previous work. But maybe WA is wrong (by the way, it happens sometimes!).
– Michael Rozenberg
Oct 21 '17 at 15:02
1
1
Delete please your "solution". I think it's not good, which you are doing.
– Michael Rozenberg
Oct 21 '17 at 16:59
Delete please your "solution". I think it's not good, which you are doing.
– Michael Rozenberg
Oct 21 '17 at 16:59
4
4
I think it's nothing.
– Michael Rozenberg
Oct 25 '17 at 11:15
I think it's nothing.
– Michael Rozenberg
Oct 25 '17 at 11:15
|
show 3 more comments
I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5a^2}geq frac{a+b}{18}$$
Proof:
We have with $x=frac{a}{b}$ :
$$frac{x^3}{13x^2+5}+frac{1}{13+5x^2}geq frac{1+x}{18}$$
Or
$$5(x+1)(x-1)^2(5x^2-8x+5)geq 0$$
So we have (if we permute the variables $a,b,c$ and addition the three inequalities ) :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}+sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{9}$$
If we have $sum_{cyc}frac{a^3}{13a^2+5b^2}geqsum_{cyc}frac{a^3}{13a^2+5c^2}$
We have :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$$
But also
$$frac{(a-epsilon)^3}{13(a-epsilon)^2+5b^2}+frac{(b)^3}{13(b)^2+5(c+epsilon)^2}+frac{(c+epsilon)^3}{13(c+epsilon)^2+5(a-epsilon)^2}geq frac{a+b+c}{18}$$
If we put $ageq c $ and $epsilon=a-c$
We finally obtain :
$$sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$$
So all the cases are here so it's proved !
1
If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions.
– Michael Rozenberg
Dec 7 at 15:38
Now it's right and proved .
– max8128
Dec 8 at 11:07
I can only see that you proved that, for a given triple $(a, b, c)$, $sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$ or $sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$ must hold. – The same applies to your other answer math.stackexchange.com/a/3041504/42969.
– Martin R
Dec 15 at 15:49
add a comment |
I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5a^2}geq frac{a+b}{18}$$
Proof:
We have with $x=frac{a}{b}$ :
$$frac{x^3}{13x^2+5}+frac{1}{13+5x^2}geq frac{1+x}{18}$$
Or
$$5(x+1)(x-1)^2(5x^2-8x+5)geq 0$$
So we have (if we permute the variables $a,b,c$ and addition the three inequalities ) :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}+sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{9}$$
If we have $sum_{cyc}frac{a^3}{13a^2+5b^2}geqsum_{cyc}frac{a^3}{13a^2+5c^2}$
We have :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$$
But also
$$frac{(a-epsilon)^3}{13(a-epsilon)^2+5b^2}+frac{(b)^3}{13(b)^2+5(c+epsilon)^2}+frac{(c+epsilon)^3}{13(c+epsilon)^2+5(a-epsilon)^2}geq frac{a+b+c}{18}$$
If we put $ageq c $ and $epsilon=a-c$
We finally obtain :
$$sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$$
So all the cases are here so it's proved !
1
If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions.
– Michael Rozenberg
Dec 7 at 15:38
Now it's right and proved .
– max8128
Dec 8 at 11:07
I can only see that you proved that, for a given triple $(a, b, c)$, $sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$ or $sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$ must hold. – The same applies to your other answer math.stackexchange.com/a/3041504/42969.
– Martin R
Dec 15 at 15:49
add a comment |
I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5a^2}geq frac{a+b}{18}$$
Proof:
We have with $x=frac{a}{b}$ :
$$frac{x^3}{13x^2+5}+frac{1}{13+5x^2}geq frac{1+x}{18}$$
Or
$$5(x+1)(x-1)^2(5x^2-8x+5)geq 0$$
So we have (if we permute the variables $a,b,c$ and addition the three inequalities ) :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}+sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{9}$$
If we have $sum_{cyc}frac{a^3}{13a^2+5b^2}geqsum_{cyc}frac{a^3}{13a^2+5c^2}$
We have :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$$
But also
$$frac{(a-epsilon)^3}{13(a-epsilon)^2+5b^2}+frac{(b)^3}{13(b)^2+5(c+epsilon)^2}+frac{(c+epsilon)^3}{13(c+epsilon)^2+5(a-epsilon)^2}geq frac{a+b+c}{18}$$
If we put $ageq c $ and $epsilon=a-c$
We finally obtain :
$$sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$$
So all the cases are here so it's proved !
I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5a^2}geq frac{a+b}{18}$$
Proof:
We have with $x=frac{a}{b}$ :
$$frac{x^3}{13x^2+5}+frac{1}{13+5x^2}geq frac{1+x}{18}$$
Or
$$5(x+1)(x-1)^2(5x^2-8x+5)geq 0$$
So we have (if we permute the variables $a,b,c$ and addition the three inequalities ) :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}+sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{9}$$
If we have $sum_{cyc}frac{a^3}{13a^2+5b^2}geqsum_{cyc}frac{a^3}{13a^2+5c^2}$
We have :
$$sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$$
But also
$$frac{(a-epsilon)^3}{13(a-epsilon)^2+5b^2}+frac{(b)^3}{13(b)^2+5(c+epsilon)^2}+frac{(c+epsilon)^3}{13(c+epsilon)^2+5(a-epsilon)^2}geq frac{a+b+c}{18}$$
If we put $ageq c $ and $epsilon=a-c$
We finally obtain :
$$sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$$
So all the cases are here so it's proved !
edited Dec 8 at 11:07
answered Dec 7 at 14:00
max8128
1,082421
1,082421
1
If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions.
– Michael Rozenberg
Dec 7 at 15:38
Now it's right and proved .
– max8128
Dec 8 at 11:07
I can only see that you proved that, for a given triple $(a, b, c)$, $sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$ or $sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$ must hold. – The same applies to your other answer math.stackexchange.com/a/3041504/42969.
– Martin R
Dec 15 at 15:49
add a comment |
1
If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions.
– Michael Rozenberg
Dec 7 at 15:38
Now it's right and proved .
– max8128
Dec 8 at 11:07
I can only see that you proved that, for a given triple $(a, b, c)$, $sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$ or $sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$ must hold. – The same applies to your other answer math.stackexchange.com/a/3041504/42969.
– Martin R
Dec 15 at 15:49
1
1
If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions.
– Michael Rozenberg
Dec 7 at 15:38
If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions.
– Michael Rozenberg
Dec 7 at 15:38
Now it's right and proved .
– max8128
Dec 8 at 11:07
Now it's right and proved .
– max8128
Dec 8 at 11:07
I can only see that you proved that, for a given triple $(a, b, c)$, $sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$ or $sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$ must hold. – The same applies to your other answer math.stackexchange.com/a/3041504/42969.
– Martin R
Dec 15 at 15:49
I can only see that you proved that, for a given triple $(a, b, c)$, $sum_{cyc}frac{a^3}{13a^2+5b^2}geq frac{a+b+c}{18}$ or $sum_{cyc}frac{a^3}{13a^2+5c^2}geq frac{a+b+c}{18}$ must hold. – The same applies to your other answer math.stackexchange.com/a/3041504/42969.
– Martin R
Dec 15 at 15:49
add a comment |
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2
Is the function $f : (a,b,c) rightarrow sumlimits_{cyc} frac{6a^3}{13a^2+5b^2} $ concave? Also, C-S = "Cauchy-Schwarz inequality"? Avoid this kind of unconventional abbreviation, please.
– Vincent
May 8 '16 at 18:17
See also math.stackexchange.com/questions/1775572/…
– David Quinn
May 8 '16 at 18:42
1
@Santropedro There is the following known inequality $sumlimits_{cyc}frac{a^3}{a^2+b^2}geqfrac{a+b+c}{2}$. I think, it's just interesting, for which maximal value of $k$ the following inequality is still true. $sumlimits_{cyc}frac{a^3}{ka^2+b^2}geqfrac{a+b+c}{k+1}$.
– Michael Rozenberg
Apr 19 '17 at 17:28
1
@Santropedro The value $,k=2.6,$ in the OP is pretty sharp. I found that $k=2.603279$ is already beyond: Then Michael's given $(0.785, 1.25, 1.861)$ violates the ineq..
– Hanno
May 11 '17 at 18:24
1
Further result obtained via random-based simulation: $(0.785, 1.2535, 1.873)$ doesn't satisfy the inequality if $,k=2.603262,$. Below that $k$-value I didn't detect a violation.
– Hanno
May 11 '17 at 18:50