Inequality – Elementary Proof for $\sqrt{xyz} \leq \dfrac{x+y+z}{3}$

Question

I am searching for an elementary proof of the AM-GM inequality in three variables:

$\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$

The inequality of the geometric mean vs the arithmetic mean of two variables can be proven elementarily via

$x – 2 \sqrt{xy} + y \geq 0$

whenever $x,y > 0$. I am searching for a proof that uses a similar technique based on elementary arithmetic.

Answer 1

Let $A=\dfrac{x+y+z}{3}$ and $G=\sqrt[3]{xyz}$

Applying AM-GM inequality for the $4$ (you can easily prove it for $4=2^2$ quantities, in general for $2^n$ quantities) quantities , $x,y,z,A$; we have $$\sqrt[4]{xyzA} \leq \dfrac{x+y+z+A}{4}$$ $$\sqrt[4]{A} \cdot G^{\frac34} \leq \dfrac{3A+A}{4}$$ $$A^{\frac14} \cdot G^{\frac34} \leq A$$ $$G^{\frac34} \leq A^{\frac34}$$ $$G \leq A$$ $$\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$$

Hope this helps.