$x,y,z>0$, $x+y+z=1$. Prove $x^2+y^2+z^2+3xyz \geq \frac{4}{9}$ 
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: 
$$2\sum_{x,y,z}(x^3+y^3-xy^2-x^2y) = 2\sum_{x,y,z}(x-y)^2(x+y)\geq 0.$$
