# Prove for the inequality between arithmetic and quadratic means $\frac{x+y}2\leq \sqrt{\frac{x^2+y^2}2}$

Show that for $x,y\in\mathbb{R}$ with $x,y\geq 0$, the arithmetic mean-quadratic mean inequality $$\frac{x+y}{2}\leq \sqrt{\frac{x^2+y^2}{2}}$$ holds.

After my calculations I'll get:

$$-x^2+2xy-y^2$$ which can't be $\leq 0$.

## 2 Answers

Completing the square gives:

$-x^2+2xy-y^2=-(x+y)^2\leq 0$

We can square both sides of your inequality, to get

$$\frac{x^2+2xy+y^2}{4} \leq \frac{x^2+y^2}{2}$$

$$\frac{x^2+2xy+y^2}{4} - \frac{x^2+y^2}{4} \leq \frac{x^2+y^2}{2}-\frac{x^2+y^2}{4}$$

$$\frac{xy}{2} \leq \frac{x^2+y^2}{4}$$

$$2xy \leq x^2+y^2$$

$$0 \leq (x-y)^2$$

This inequality is true, since $x-y$ is a real number and all squares of real numbers are non-negative. We have shown that your inequality is equivalent to one that we know is always true. Therefore yours is always true.

• It is not true that $a^2\geq a$ for all real $a$, but it doesn't matter. We can still justify squaring each side of the inequality. If the LHS is negative then the inequality obviously holds. Otherwise, each side is non-negative, and $f (x)=x^2$ is an increasing function for non-negative $x$. – Dylan Nov 2 '15 at 7:50
• @Dylan Thanks for the correction - not sure why I wrote that, seems very silly in hindsight. – Zubin Mukerjee Nov 2 '15 at 9:41