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

Question

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 $\leq 0$.

Answer 1

Completing the square gives:

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

Answer 2

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.