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 $\leq 0$.

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. 
