Prove $\frac{1}{3}(x+y+z)^2 \geq xy + yz + xz.$ 
Prove that for nonnegative $x,y,z$ that $\frac{1}{3}(x+y+z)^2 \geq xy + yz + xz.$

I saw this result in a problem but didn't know how to prove it. I tried expanding and collecting to get the trivial inequality but it didn't work.
The reason I am asking this is because it comes from the solution below where it claims it is true.

We note that $S_a=ad_a/2$, $S_b=bd_b/2$, and $S_c=cd_c/2$ are the areas of the triangles $MBC$, $MCA$, and $MAB$ respectively. The desired inequality now follows from $$S_aS_b+S_bS_c+S_cS_a \le \frac{1}{3}(S_a+S_b+S_c)^2=\frac{S^2}{3}.$$ Equality holds if and only if $S_a=S_b=S_c$, which is equivalent to $M$ being the center of the triangle.


$\frac{1}{3}(x+y+z)^2 = \frac{2}{3}(xy + zx + yz) + \frac{1}{3}(y^2 + z^2 + x^2) \geq xy + yz + xz.$
since per Cauchy Schwarz we have $(y^2 + z^2 + x^2)^2\geq (yz + xy + xz)^2$ which implies $(y^2 + z^2 + x^2) \geq (yz + xy + xz)$ since square-rooting preserves order assuming a positive argument.
