Prove that $\sqrt[4]{xyzw} \leq \frac{x+y+z+w}{4}$ for any $x, y, z, w \geq 0$

Question

Prove that

$\sqrt[4]{xyzw} \leq \frac{x+y+z+w}{4}$

For any $x, y, z, \geq 0$

And prove the AM-GM inequality for three numbers, $\sqrt[3]{xyz} \leq \frac{x+y+z}{3}$ where $x, y, z \geq 0$, by using the previous proof with $w= (xyz)^{1/3}$

Answer 1

You are on the good track, indeed let consider by $AM-GM$

$$u=\frac{x+y}{2}\ge \sqrt{xy} \quad v=\frac{z+w}{2}\ge \sqrt{zw}$$

then

$$\frac{u+v}{2}= \frac{x+y+z+w}{4}\ge \sqrt{uv}\ge \sqrt{\sqrt{xy}\sqrt{zw}}=\sqrt[4]{xyzw}$$