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Root inequality

Let a, b, c be positive real numbers such that a+b+c=3. Prove that

    \[\sqrt{\frac{b}{a^2+3}} + \sqrt{\frac{c}{b^2+3}} + \sqrt{\frac{a}{c^2+3}} \leq \frac{3}{2} \sqrt[4]{\frac{1}{abc}}\]

Solution

Due to the AM – GM we have that

(1)   \begin{equation*} a^2+3 \geq 4\sqrt{a} \end{equation*}

and

(2)   \begin{equation*} 3b + c \geq 4 \sqrt[4]{b^3c} \end{equation*}

Thus,

    \begin{align*} \sum \sqrt{\frac{b}{a^2+3}}  &\leq \sqrt{\frac{b}{4\sqrt{a}}} + \sqrt{\frac{c}{4\sqrt{b}}} + \sqrt{\frac{a}{4\sqrt{c}}} \\ &=\frac{1}{2\sqrt[4]{abc}} \left ( \sqrt[4]{b^3c} + \sqrt[4]{c^3a} + \sqrt[4]{a^3b} \right ) \\ &\leq \frac{1}{2\sqrt[4]{abc}} \left ( \frac{3b+c}{4} + \frac{3c+a}{4} + \frac{3a+b}{4} \right ) \\ &=\frac{3}{2} \sqrt[4]{\frac{1}{abc}} \end{align*}

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