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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

Well if we apply AM-GM to (a^2, 1, 1, 1) we obtain

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

and similarly if we apply AM – GM to (b, b, b, c) we obtain

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

We have successively,

\begin{aligned} \sqrt{\frac{b}{a^2+3}} + \sqrt{\frac{c}{b^2+3}} + \sqrt{\frac{a}{c^2+3}} &\overset{(1)}{\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 ) \\ &\overset{(2)}{\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{aligned}

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