Generalization of Cauchy Inequality

Algebra Level 4

a 2017 b + c + b 2017 c + a + c 2017 a + b a k + b k + c k 2 \large\ \frac { { a }^{ 2017 } }{ b+c } +\frac { { b }^{ 2017 } }{ c+a } +\frac { { c }^{ 2017 } }{ a+b } \ge \frac { { a }^{ k }+{ b }^{ k }+{ c }^{ k } }{ 2 }

If this inequality is satisfied for one value of k k , find k k .


The answer is 2016.

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1 solution

Aaron Jerry Ninan
Sep 24, 2016

By applying Chebeshev's inequality 3 ( a b + c a 2016 + b c + a b 2016 + c a + b c 2016 ) ( a b + c + b c + a + c a + b ) ( a 2016 + b 2016 + c 2016 ) 3(\frac{a}{b+c}a^{2016}+\frac{b}{c+a}b^{2016}+\frac{c}{a+b}c^{2016})\geq (\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})(a^{2016}+b^{2016}+c^{2016}) But by our classic AM-HM problem ,we know that, a b + c + b c + a + c a + b 3 2 \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2} Thus we get, 3 ( a b + c a 2016 + b c + a b 2016 + c a + b c 2016 ) 3 2 ( a 2016 + b 2016 + c 2016 ) 3(\frac{a}{b+c}a^{2016}+\frac{b}{c+a}b^{2016}+\frac{c}{a+b}c^{2016})\geq\frac{3}{2} (a^{2016}+b^{2016}+c^{2016}) Finally we get, ( a 2017 + b 2017 + c 2017 ) 1 2 ( a 2016 + b 2016 + c 2016 ) (a^{2017}+b^{2017}+c^{2017})\geq \frac{1}{2} (a^{2016}+b^{2016}+c^{2016}) Therefore K=2016

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