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.


Like Tunococ's solution but with a modification.
After expanding, we need to show,
$x^2 + y^2 + z^2 \ge xy + xz + yz$
By the Cauchy-Schwarz inequality,
$3(x^2 + y^2 + z^2) \ge (x + y + z)^2$
This means,
$2(x^2 + y^2 + z^2 - xy - yz - xz) \ge 0 \implies x^2 + y^2 + z^2 \ge xy + yz + xz$.
