Question

The minimum value of $\frac{x^4+y^4+z^2}{xyz}$ for positive real numbers $x,y,z$ is

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

Answer 1

The correct option is B $2\sqrt{2}$

we know that $A.M.\ge G.M.$

$⟹\frac{x^4+y^4+\frac{z^2}{2}+\frac{z^2}{2}}{4}\ge \sqrt[4]{4x^4\times y^4\times \frac{z^2}{2}\times \frac{z^2}{2}}$

$⟹\frac{x^4+y^4+\frac{z^2}{2}+\frac{z^2}{2}}{4}\ge \sqrt[4]{4x^4\times y^4\times \frac{z^4}{4}}$

$⟹\frac{x^4+y^4+\frac{z^2}{2}+\frac{z^2}{2}}{4}\ge xyz\sqrt[4]{4\frac{1}{4}}$

$⟹\frac{x^4+y^4+\frac{2z^2}{2}}{xyz}\ge 4\sqrt[4]{4\frac{1}{4}}$

$⟹\frac{x^4+y^4+z^2}{xyz}\ge \sqrt[4]{4\frac{4^4}{4}}$

$⟹\frac{x^4+y^4+z^2}{xyz}\ge \sqrt[4]{44^3}$

$⟹\frac{x^4+y^4+z^2}{xyz}\ge \sqrt[4]{42^6}$

$⟹\frac{x^4+y^4+z^2}{xyz}\ge 2^{\frac{6}{4}}$

$⟹\frac{x^4+y^4+z^2}{xyz}\ge 2^{\frac{3}{2}}$

$⟹\frac{x^4+y^4+z^2}{xyz}\ge \sqrt{8}$

$⟹\frac{x^4+y^4+z^2}{xyz}\ge 2\sqrt{2}$

therefore the minimum value of $\frac{x^4+y^4+z^2}{xyz}$ is $2\sqrt{2}$