If $a+b+c=frac 1a +frac 1b +frac 1c$ then prove that $ab+bc+ca geq 3$
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Let $a,b,c$ be positive real numbers. $a+b+c=frac 1a +frac 1b +frac 1c$ then prove that $ab+bc+ca geq 3$
Using CS Inequality $(a+b+c)(frac 1a +frac 1b +frac 1c) >9$. then by hyp $frac 1a +frac 1b +frac 1c >3$. Now can we prove that $abc>1$?
inequality contest-math
asked Nov 30 '18 at 4:03
GimgimGimgim
1869
$endgroup$
add a comment |
$begingroup$
Let $a,b,c$ be positive real numbers. $a+b+c=frac 1a +frac 1b +frac 1c$ then prove that $ab+bc+ca geq 3$
Using CS Inequality $(a+b+c)(frac 1a +frac 1b +frac 1c) >9$. then by hyp $frac 1a +frac 1b +frac 1c >3$. Now can we prove that $abc>1$?
inequality contest-math
asked Nov 30 '18 at 4:03
GimgimGimgim
1869
$endgroup$
$begingroup$
Yes, sorry. I will add that
$endgroup$
– Gimgim
Nov 30 '18 at 4:24
add a comment |
$begingroup$
Let $a,b,c$ be positive real numbers. $a+b+c=frac 1a +frac 1b +frac 1c$ then prove that $ab+bc+ca geq 3$
Using CS Inequality $(a+b+c)(frac 1a +frac 1b +frac 1c) >9$. then by hyp $frac 1a +frac 1b +frac 1c >3$. Now can we prove that $abc>1$?
inequality contest-math
asked Nov 30 '18 at 4:03
GimgimGimgim
1869
$endgroup$
Let $a,b,c$ be positive real numbers. $a+b+c=frac 1a +frac 1b +frac 1c$ then prove that $ab+bc+ca geq 3$
Using CS Inequality $(a+b+c)(frac 1a +frac 1b +frac 1c) >9$. then by hyp $frac 1a +frac 1b +frac 1c >3$. Now can we prove that $abc>1$?
inequality contest-math
inequality contest-math
asked Nov 30 '18 at 4:03
GimgimGimgim
1869
asked Nov 30 '18 at 4:03
GimgimGimgim
1869
edited Nov 30 '18 at 4:25
Gimgim
asked Nov 30 '18 at 4:03
GimgimGimgim
1869
asked Nov 30 '18 at 4:03
GimgimGimgim
1869
asked Nov 30 '18 at 4:03
GimgimGimgim
1869
1869
$begingroup$
Yes, sorry. I will add that
$endgroup$
– Gimgim
Nov 30 '18 at 4:24
add a comment |
$begingroup$
Yes, sorry. I will add that
$endgroup$
– Gimgim
Nov 30 '18 at 4:24
$begingroup$
Yes, sorry. I will add that
$endgroup$
– Gimgim
Nov 30 '18 at 4:24
$begingroup$
Yes, sorry. I will add that
$endgroup$
– Gimgim
Nov 30 '18 at 4:24
add a comment |
1 Answer
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$begingroup$
The condition gives $$1=frac{abc(a+b+c)}{ab+ac+bc}$$
Id est, we need to prove that
$$ab+ac+bcgeqfrac{3abc(a+b+c)}{ab+ac+bc}$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c)$$ or
$$sum_{cyc}(a^2b^2+2a^2bc)geq3sum_{cyc}a^2bc$$ or
$$sum_{cyc}c^2(a-b)^2geq0.$$
If you want to prove that $abcgeq1$ then it's impossible because it's wrong.
Indeed, $$abcgeq1$$ it's
$$abcgeqleft(sqrt{frac{abc(a+b+c)}{ab+ac+bc}}right)^3$$ or
$$ab+ac+bcgeq(a+b+c)sqrt[3]{abc},$$ which is wrong for $arightarrow+infty.$
answered Nov 30 '18 at 4:31
Michael RozenbergMichael Rozenberg
98.3k1590188
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
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active
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$begingroup$
The condition gives $$1=frac{abc(a+b+c)}{ab+ac+bc}$$
Id est, we need to prove that
$$ab+ac+bcgeqfrac{3abc(a+b+c)}{ab+ac+bc}$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c)$$ or
$$sum_{cyc}(a^2b^2+2a^2bc)geq3sum_{cyc}a^2bc$$ or
$$sum_{cyc}c^2(a-b)^2geq0.$$
If you want to prove that $abcgeq1$ then it's impossible because it's wrong.
Indeed, $$abcgeq1$$ it's
$$abcgeqleft(sqrt{frac{abc(a+b+c)}{ab+ac+bc}}right)^3$$ or
$$ab+ac+bcgeq(a+b+c)sqrt[3]{abc},$$ which is wrong for $arightarrow+infty.$
answered Nov 30 '18 at 4:31
Michael RozenbergMichael Rozenberg
98.3k1590188
$endgroup$
add a comment |
$begingroup$
The condition gives $$1=frac{abc(a+b+c)}{ab+ac+bc}$$
Id est, we need to prove that
$$ab+ac+bcgeqfrac{3abc(a+b+c)}{ab+ac+bc}$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c)$$ or
$$sum_{cyc}(a^2b^2+2a^2bc)geq3sum_{cyc}a^2bc$$ or
$$sum_{cyc}c^2(a-b)^2geq0.$$
If you want to prove that $abcgeq1$ then it's impossible because it's wrong.
Indeed, $$abcgeq1$$ it's
$$abcgeqleft(sqrt{frac{abc(a+b+c)}{ab+ac+bc}}right)^3$$ or
$$ab+ac+bcgeq(a+b+c)sqrt[3]{abc},$$ which is wrong for $arightarrow+infty.$
answered Nov 30 '18 at 4:31
Michael RozenbergMichael Rozenberg
98.3k1590188
$endgroup$
add a comment |
$begingroup$
The condition gives $$1=frac{abc(a+b+c)}{ab+ac+bc}$$
Id est, we need to prove that
$$ab+ac+bcgeqfrac{3abc(a+b+c)}{ab+ac+bc}$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c)$$ or
$$sum_{cyc}(a^2b^2+2a^2bc)geq3sum_{cyc}a^2bc$$ or
$$sum_{cyc}c^2(a-b)^2geq0.$$
If you want to prove that $abcgeq1$ then it's impossible because it's wrong.
Indeed, $$abcgeq1$$ it's
$$abcgeqleft(sqrt{frac{abc(a+b+c)}{ab+ac+bc}}right)^3$$ or
$$ab+ac+bcgeq(a+b+c)sqrt[3]{abc},$$ which is wrong for $arightarrow+infty.$
answered Nov 30 '18 at 4:31
Michael RozenbergMichael Rozenberg
98.3k1590188
$endgroup$
The condition gives $$1=frac{abc(a+b+c)}{ab+ac+bc}$$
Id est, we need to prove that
$$ab+ac+bcgeqfrac{3abc(a+b+c)}{ab+ac+bc}$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c)$$ or
$$sum_{cyc}(a^2b^2+2a^2bc)geq3sum_{cyc}a^2bc$$ or
$$sum_{cyc}c^2(a-b)^2geq0.$$
If you want to prove that $abcgeq1$ then it's impossible because it's wrong.
Indeed, $$abcgeq1$$ it's
$$abcgeqleft(sqrt{frac{abc(a+b+c)}{ab+ac+bc}}right)^3$$ or
$$ab+ac+bcgeq(a+b+c)sqrt[3]{abc},$$ which is wrong for $arightarrow+infty.$
answered Nov 30 '18 at 4:31
Michael RozenbergMichael Rozenberg
98.3k1590188
edited Nov 30 '18 at 4:39
answered Nov 30 '18 at 4:31
Michael RozenbergMichael Rozenberg
98.3k1590188
answered Nov 30 '18 at 4:31
Michael RozenbergMichael Rozenberg
98.3k1590188
answered Nov 30 '18 at 4:31
Michael RozenbergMichael Rozenberg
98.3k1590188
98.3k1590188
add a comment |
add a comment |
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$begingroup$
Yes, sorry. I will add that
$endgroup$
– Gimgim
Nov 30 '18 at 4:24