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. $\ge$ G.M.
$\Rightarrow \frac{x^4+y^4+\frac{z^2}{2}+\frac{z^2}{2}}{4}\ge \sqrt[4]{x^4\times y^4\times \frac{z^2}{2}\times \frac{z^2}{2}}$
$\Rightarrow \frac{x^4+y^4+\frac{z^2}{2}+\frac{z^2}{2}}{4}\ge \sqrt[4]{x^4\times y^4\times \frac{z^4}{4}}$
$\Rightarrow \frac{x^4+y^4+\frac{z^2}{2}+\frac{z^2}{2}}{4}\ge xyz\sqrt[4]{\frac{1}{4}}$
$\Rightarrow \frac{x^4+y^4+\frac{2z^2}{2}}{xyz}\ge 4\sqrt[4]{\frac{1}{4}}$
$\Rightarrow \frac{x^4+y^4+z^2}{xyz}\ge \sqrt[4]{\frac{4^4}{4}}$
$\Rightarrow \frac{x^4+y^4+z^2}{xyz}\ge \sqrt[4]{4^3}$
$\Rightarrow \frac{x^4+y^4+z^2}{xyz}\ge \sqrt[4]{2^6}$
$\Rightarrow \frac{x^4+y^4+z^2}{xyz}\ge 2^{\frac{6}{4}}$
$\Rightarrow \frac{x^4+y^4+z^2}{xyz}\ge 2^{\frac{3}{2}}$
$\Rightarrow \frac{x^4+y^4+z^2}{xyz}\ge \sqrt{8}$
$\Rightarrow \frac{x^4+y^4+z^2}{xyz}\ge 2\sqrt{2}$
therefore the minimum value of $<br/>\frac{x^4+y^4+z^2}{xyz}$ is $2\sqrt{2}$