Question

The minimum value of $\frac{x^{4} +y^{4} +z^{4}}{xyz} $ for positive real number $x, y, z$ is

• A. $\sqrt{2}$
• B. $ 2 \sqrt{2}$
• C. $ 4 \sqrt{2}$
• D. $ 8 \sqrt{2}$

Answer 1

Answer: (B)

we know that A.M. $\geq$ G.M.
$\Rightarrow \frac{x^{4}+y^{4}+\frac{z^{2}}{2}+\frac{z^{2}}{2}}{4} \geq \sqrt[4]{x^{4} \times y^{4} \times \frac{z^{2}}{2} \times \frac{z^{2}}{2}}$
$\Rightarrow \frac{x^{4}+y^{4}+\frac{z^{2}}{2}+\frac{z^{2}}{2}}{4} \geq \sqrt[4]{x^{4} \times y^{4} \times \frac{z^{4}}{4}}$
$\Rightarrow \frac{x^{4}+y^{4}+\frac{z^{2}}{2}+\frac{z^{2}}{2}}{4} \geq x y z \sqrt[4]{\frac{1}{4}}$
$\Rightarrow \frac{x^{4}+y^{4}+\frac{2 z^{2}}{2}}{x y z} \geq 4 \sqrt[4]{\frac{1}{4}}$
$\Rightarrow \frac{x^{4}+y^{4}+z^{2}}{x y z} \geq \sqrt[4]{\frac{4^{4}}{4}}$
$\Rightarrow \frac{x^{4}+y^{4}+z^{2}}{x y z} \geq \sqrt[4]{4^{3}}$
$\Rightarrow \frac{x^{4}+y^{4}+z^{2}}{x y z} \geq \sqrt[4]{2^{6}}$
$\Rightarrow \frac{x^{4}+y^{4}+z^{2}}{x y z} \geq 2^{\frac{6}{4}}$
$\Rightarrow \frac{x^{4}+y^{4}+z^{2}}{x y z} \geq 2^{\frac{3}{2}}$
$\Rightarrow \frac{x^{4}+y^{4}+z^{2}}{x y z} \geq \sqrt{8}$
$\Rightarrow \frac{x^{4}+y^{4}+z^{2}}{x y z} \geq 2 \sqrt{2}$
therefore the minimum value of $
\frac{x^{4}+y^{4}+z^{2}}{x y z}$ is $2 \sqrt{2}$