Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that 
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...

$\frac12 (x^2+y^2)-(\frac12 (x+y))^2$
$=\frac12 (x^2+y^2)-\frac14 (x^2+y^2+2xy)$
$=\frac14 (2x^2+2y^2-x^2-y^2-2xy)$
$=\frac14 (x^2+y^2-2xy)$
$=\frac14 (x-y)^2\ge 0$
