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...

This substitution is a nice way to do this:
$$x=r\sin\theta , y=r\cos\theta$$
So, our identity:
$$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
$$\frac{1}{4} (x^2+y^2+2xy) \leq \frac12(x^2 + y^2)$$
Applying our substitutions:
$$\frac{1}{4} (r^2+2\cdot r\sin\theta\cdot r\cos\theta) \leq \frac12(r^2)$$
$$\frac{r^2}{2}\cdot \frac{1+\sin2\theta}{2} \leq \frac{r^2}{2}$$
Now, if you note that the maximum value of $\sin 2\theta$ is $1$, then it becomes obvious.
Added: It could have also been done it this way:
$$\left(\frac{1}{2}(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
$$\frac{1}{2}\cdot\frac{1}{2}(r\cos\theta+r\sin\theta)^2 \leq \frac{r^2}{2}$$
$$\frac{r^2}{2}\cdot\frac{1}{2}(\cos\theta+\sin\theta)^2 \leq \frac{r^2}{2}$$
Now, note that $-\sqrt{2} \leq \cos\theta+\sin\theta = \sqrt{2}\cos(x+45^{\circ}) \leq \sqrt{2}$. Thus the maximum value of $(\cos\theta+\sin\theta)^2$ is $2$, and the result follow.
