Let $x$, $y$, $z$ be positive real numbers such that $x+y+z=1$, then $$ x^2+y^2+z^2+3xyz \geq \frac{4}{9}.$$
From Cauchy–Schwarz inequality, we have $$ \left( x^2+y^2+z^2 \right)\left(1^2+1^2+1^2\right)\geq(x+y+z)^2 = 1 \implies x^2+y^2+z^2 \geq \frac{1}{3}. $$
It suffices to prove $$ xyz \geq \frac{1}{27}. $$
However by AM–GM inequality,
$$ \frac{1}{3} = \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \implies xyz \leq \frac{1}{27}. $$
Am I missing something?
As commented by @mfl, we have
$$ \begin{aligned} &9(x+y+z)\left(x^2+y^2+z^2\right) + 27xyz \geq 4(x+y+z)^3 \\ \iff &5\left(x^3+y^3+z^3\right) + 3xyz \geq 3\left(x^2(y+z)+y^2(x+z)+z^2(x+y)\right) \\ \iff &4\left(x^3+y^3+z^3\right) \geq 2\left(x^2(y+z)+y^2(x+z)+z^2(x+y)\right) \end{aligned} $$
by Schur’s inequality. How to go further from this?
Rewrite your last line as: $$2\sum_{x,y,z}(x^3+y^3-xy^2-x^2y) = 2\sum_{x,y,z}(x-y)^2(x+y)\geq 0.$$