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.


Here's my attempt to expand and collect terms:
\begin{align}
\frac 13(x + y + z)^2 & \stackrel{?}{\ge} xy + xz + yz\\
x^2 + y^2 + z^2 + 2xy + 2xz + 2yz & \stackrel{?}{\ge} 3xy + 3xz + 3yz \\
x^2 + y^2 + z^2 - xy - xz - yz & \stackrel{?}{\ge} 0 \\
\frac 12(x - y)^2 + \frac 12(x - z)^2 + \frac 12(y - z)^2 & \stackrel{?}{\ge} 0.
\end{align}
