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Root inequality

Let a, b, c be three positive real numbers such that \sqrt{a} + \sqrt{b} + \sqrt{c}=1. Prove that

    \[\frac{\sqrt{a}}{a^2+2bc} + \frac{\sqrt{b}}{b^2+2ca} + \frac{\sqrt{c}}{c^2+2ab} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\]

Solution

By AM – GM we have,

    \begin{align*} \sum \frac{\sqrt{a}}{a^2+2bc} &\leq \sum \frac{\sqrt{a}}{a^2 + 2\left ( \frac{b^2+c^2}{2} \right )} \\ &= \frac{1}{a^2+b^2+c^2} \sum \sqrt{a}\\ &= \frac{1}{a^2+b^2+c^2} \end{align*}

However,

    \begin{align*} 1 &= \left (\sum \sqrt{a} \right )^2 \\ &=\left ( \sum \frac{a}{\sqrt{a}} \right )^2 \\ &\leq \left ( \sum a^2 \right ) \cdot \left ( \sum \frac{1}{a} \right ) \end{align*}

Hence \displaystyle \frac{1}{a^2+b^2+c^2} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} and the exercise is complete.

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