Problem 21: Hard inequality

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by henderson, Jun 8, 2016, 2:34 PM

$\bf\color{red}Problem\ 21\$ Let $a,b,c$ be positive reals such that ${a^2} + {b^2} + {c^2} = 3$ . Prove that
$2\left( {a + b + c} \right) + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge 9.$ $\bf\color{green}Solution\$ Because $a^2 < 3 < 4,$ $0 < a < 2$ is true. So,
$2a+\frac{1}{a}-\left (3+\frac{1}{2}(a^2-1) \right ) = \frac{(2-a)(a-1)^2}{2a} \geq 0.$ Therefore
$\sum_{cyc} \left (2a+\frac{1}{a} \right ) \geq \sum_{cyc} \left (3+\frac{1}{2}(a^2-1) \right ) = 9.$ The proof is completed.

This post has been edited 5 times. Last edited by henderson, Sep 11, 2016, 7:46 PM

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Problem 21: Hard inequality

by henderson, Jun 8, 2016, 2:34 PM

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