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

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 $w= (xyz)^{1/3}$

• Are you familiar with the AM-GM inequality? – user170231 Sep 18 '18 at 14:50
• Yes, I am, however when I try $\sqrt{\sqrt{xy}\sqrt{zw}} \leq \frac{\sqrt{xy}+\sqrt{zw}}{2}$ I'm unsure if I'm correct in proceeding as such, and if so, I'm unsure how to proceed from this point onward. – Kyle Xiao Sep 18 '18 at 14:56
• @KyleXiao, what information do you assume? For example, the AM - GM inequality for two numbers, etc. – AnotherJohnDoe Sep 18 '18 at 15:02
• What you write in your comment is correct. To proceed onward, what can you say about $\sqrt{xy}$? – Arnaud D. Sep 18 '18 at 15:02
• Well $\sqrt{xy} \leq \frac{x+y}{2}$ but would I be right in using $\sqrt{xy} = \frac{x+y}{2}$ to substitute – Kyle Xiao Sep 18 '18 at 15:03

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

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

Hint: The AM-GM inequality can be proven for $n$ variables in the following way:

1. Show it for $2$ variables
2. Prove by induction that if you know it for $2^n$ variables, you can show it for $2^{n+1}$ variables by first applying it on $x_1,\cdots,x_{2^n}$, then on $x_{2^n+1},\cdots,x_{2^{n+1}}$ - then use the $2$-variable version on the results of those two applications
3. Prove that if you know it for $N+1$ you know it for $N$ by applying it on $x_1,\cdots,x_N,\frac{x_1+x_2+\cdots+x_N}{N}$.

Many of these steps will require some algebraic manipulation.