Inequality for $a,c,b$ positive real numbers with $a + b + c = 1$
2
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Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$frac{1}{2a^2+bc}+frac{1}{2b^2+ac}+frac{1}{2c^2+ab}geq frac{1}{sqrt{a^2-ab+b^2}}+frac{1}{sqrt{b^2-bc+c^2}}+frac{1}{sqrt{c^2-ca+a^2}}$$
I have a solution enter image description here. I need another solution. Thanks all
inequality
edited Dec 3 '18 at 13:57
Mariah
1,4621518
asked Dec 3 '18 at 13:29
Winter In My Heart Winter In My Heart
463
$endgroup$
add a comment |
2
$begingroup$
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$frac{1}{2a^2+bc}+frac{1}{2b^2+ac}+frac{1}{2c^2+ab}geq frac{1}{sqrt{a^2-ab+b^2}}+frac{1}{sqrt{b^2-bc+c^2}}+frac{1}{sqrt{c^2-ca+a^2}}$$
I have a solution enter image description here. I need another solution. Thanks all
inequality
edited Dec 3 '18 at 13:57
Mariah
1,4621518
asked Dec 3 '18 at 13:29
Winter In My Heart Winter In My Heart
463
$endgroup$
1
$begingroup$
In My Heart Your solution is very nice! There is also solution by C-S. If in evening your topic not will be closed, I'll try to show my proof.
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 14:55
$begingroup$
Why do you need another solution? Are you worried you have made a mistake in your solution?
$endgroup$
– Robert Soupe
Dec 3 '18 at 15:31
1
$begingroup$
@Winter In My Heart C-S with $(b+c)^2$ in the LHS and C-S with $sumlimits_{cyc}frac{a+b}{a^2-ab+b^2}$ in the RHS gives a right inequality, but your solution is much more easier and now I don't want to post my solution. Your proof is extremely beautiful. My congratulations!
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 17:54
$begingroup$
@MichaelRozenberg Show me your proof, please. By C-S or uvw or Vasc LCF...
$endgroup$
– Winter In My Heart
Dec 3 '18 at 22:32
add a comment |
2
2
2
2
$begingroup$
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$frac{1}{2a^2+bc}+frac{1}{2b^2+ac}+frac{1}{2c^2+ab}geq frac{1}{sqrt{a^2-ab+b^2}}+frac{1}{sqrt{b^2-bc+c^2}}+frac{1}{sqrt{c^2-ca+a^2}}$$
I have a solution enter image description here. I need another solution. Thanks all
inequality
edited Dec 3 '18 at 13:57
Mariah
1,4621518
asked Dec 3 '18 at 13:29
Winter In My Heart Winter In My Heart
463
$endgroup$
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$frac{1}{2a^2+bc}+frac{1}{2b^2+ac}+frac{1}{2c^2+ab}geq frac{1}{sqrt{a^2-ab+b^2}}+frac{1}{sqrt{b^2-bc+c^2}}+frac{1}{sqrt{c^2-ca+a^2}}$$
I have a solution enter image description here. I need another solution. Thanks all
inequality
inequality
edited Dec 3 '18 at 13:57
Mariah
1,4621518
asked Dec 3 '18 at 13:29
Winter In My Heart Winter In My Heart
463
edited Dec 3 '18 at 13:57
Mariah
1,4621518
asked Dec 3 '18 at 13:29
Winter In My Heart Winter In My Heart
463
edited Dec 3 '18 at 13:57
Mariah
1,4621518
edited Dec 3 '18 at 13:57
Mariah
1,4621518
edited Dec 3 '18 at 13:57
Mariah
1,4621518
1,4621518
asked Dec 3 '18 at 13:29
Winter In My Heart Winter In My Heart
463
asked Dec 3 '18 at 13:29
Winter In My Heart Winter In My Heart
463
asked Dec 3 '18 at 13:29
Winter In My Heart Winter In My Heart
463
463
1
$begingroup$
In My Heart Your solution is very nice! There is also solution by C-S. If in evening your topic not will be closed, I'll try to show my proof.
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 14:55
$begingroup$
Why do you need another solution? Are you worried you have made a mistake in your solution?
$endgroup$
– Robert Soupe
Dec 3 '18 at 15:31
1
$begingroup$
@Winter In My Heart C-S with $(b+c)^2$ in the LHS and C-S with $sumlimits_{cyc}frac{a+b}{a^2-ab+b^2}$ in the RHS gives a right inequality, but your solution is much more easier and now I don't want to post my solution. Your proof is extremely beautiful. My congratulations!
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 17:54
$begingroup$
@MichaelRozenberg Show me your proof, please. By C-S or uvw or Vasc LCF...
$endgroup$
– Winter In My Heart
Dec 3 '18 at 22:32
add a comment |
1
$begingroup$
In My Heart Your solution is very nice! There is also solution by C-S. If in evening your topic not will be closed, I'll try to show my proof.
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 14:55
$begingroup$
Why do you need another solution? Are you worried you have made a mistake in your solution?
$endgroup$
– Robert Soupe
Dec 3 '18 at 15:31
1
$begingroup$
@Winter In My Heart C-S with $(b+c)^2$ in the LHS and C-S with $sumlimits_{cyc}frac{a+b}{a^2-ab+b^2}$ in the RHS gives a right inequality, but your solution is much more easier and now I don't want to post my solution. Your proof is extremely beautiful. My congratulations!
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 17:54
$begingroup$
@MichaelRozenberg Show me your proof, please. By C-S or uvw or Vasc LCF...
$endgroup$
– Winter In My Heart
Dec 3 '18 at 22:32
1
1
$begingroup$
In My Heart Your solution is very nice! There is also solution by C-S. If in evening your topic not will be closed, I'll try to show my proof.
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 14:55
$begingroup$
In My Heart Your solution is very nice! There is also solution by C-S. If in evening your topic not will be closed, I'll try to show my proof.
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 14:55
$begingroup$
Why do you need another solution? Are you worried you have made a mistake in your solution?
$endgroup$
– Robert Soupe
Dec 3 '18 at 15:31
$begingroup$
Why do you need another solution? Are you worried you have made a mistake in your solution?
$endgroup$
– Robert Soupe
Dec 3 '18 at 15:31
1
1
$begingroup$
@Winter In My Heart C-S with $(b+c)^2$ in the LHS and C-S with $sumlimits_{cyc}frac{a+b}{a^2-ab+b^2}$ in the RHS gives a right inequality, but your solution is much more easier and now I don't want to post my solution. Your proof is extremely beautiful. My congratulations!
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 17:54
$begingroup$
@Winter In My Heart C-S with $(b+c)^2$ in the LHS and C-S with $sumlimits_{cyc}frac{a+b}{a^2-ab+b^2}$ in the RHS gives a right inequality, but your solution is much more easier and now I don't want to post my solution. Your proof is extremely beautiful. My congratulations!
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 17:54
$begingroup$
@MichaelRozenberg Show me your proof, please. By C-S or uvw or Vasc LCF...
$endgroup$
– Winter In My Heart
Dec 3 '18 at 22:32
$begingroup$
@MichaelRozenberg Show me your proof, please. By C-S or uvw or Vasc LCF...
$endgroup$
– Winter In My Heart
Dec 3 '18 at 22:32
add a comment |
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1
$begingroup$
In My Heart Your solution is very nice! There is also solution by C-S. If in evening your topic not will be closed, I'll try to show my proof.
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 14:55
$begingroup$
Why do you need another solution? Are you worried you have made a mistake in your solution?
$endgroup$
– Robert Soupe
Dec 3 '18 at 15:31
1
$begingroup$
@Winter In My Heart C-S with $(b+c)^2$ in the LHS and C-S with $sumlimits_{cyc}frac{a+b}{a^2-ab+b^2}$ in the RHS gives a right inequality, but your solution is much more easier and now I don't want to post my solution. Your proof is extremely beautiful. My congratulations!
$endgroup$
– Michael Rozenberg
Dec 3 '18 at 17:54
$begingroup$
@MichaelRozenberg Show me your proof, please. By C-S or uvw or Vasc LCF...
$endgroup$
– Winter In My Heart
Dec 3 '18 at 22:32