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.

 A: 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}
A: If you multiply both sides by $6$ and then expand the square, you get $$2x^2+2y^2+2z^2+4xy+4yz+4xz\geq 6xy+6yz+6xz,$$ which you can rewrite as $$(x^2+y^2)+(y^2+z^2)+(x^2+z^2)\geq 2xy+2yz+2xz.$$ Then rewrite this as $$(x-y)^2+(y-z)^2+(x-z)^2\geq 0.$$
A: The equation, after simplification, reduces to 
$$S_a^2 + S_b^2 + S_c^2 \ge S_aS_b + S_bS_c + S_cS_a$$
We know that A.M $\ge$ G.M or $${a+b\over 2} \ge \sqrt{ab} \implies {(S_a^2 + S_b^2)\over 2} \ge S_aS_b$$
Extend this to the other variables: $${(S_b^2 + S_c^2)\over 2} \ge S_bS_c \qquad {(S_c^2 + S_a^2)\over 2} \ge S_cS_a$$
Adding all the inequalities, we get $$S_a^2 + S_b^2 + S_c^2 \ge S_aS_b + S_bS_c + S_cS_a$$
A: Just so you know, with real constant $\lambda,$ the quadratic form
$$ x^2 + y^2 + z^2 + \lambda (yz + zx + xy) $$
is positive definite for $-1 < \lambda < 2.$ It is positive semidefinite for $\lambda = 2$ and for $\lambda = -1.$ 
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
Why not. Sylvester is applied to the Hessian matrix of second partial derivatives,
$$
\left(
\begin{array}{ccc}
2 & \lambda & \lambda \\
 \lambda & 2 & \lambda \\
 \lambda & \lambda & 2 \\
\end{array}
\right)
$$
Note that the eigenvalues are $\{2 + 2 \lambda, 2 - \lambda, 2 - \lambda \}$
A: 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,
$2(x^2 + y^2 + z^2 - xy - yz - xz) \ge 0 \implies x^2 + y^2 + z^2 \ge xy + yz + xz$.
A: $\frac{1}{3}(x+y+z)^2 = \frac{2}{3}(xy + zx + yz) + \frac{1}{3}(y^2 + z^2 + x^2) \geq xy + yz + xz.$
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.
