How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$? How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$?

You can write the inequality as 
$$
xyz(x+y+z)\leq\frac13. 
$$
On the left-hand-side, using Cauchy-Schwarz, 
$$
xyz(x+y+z)\leq xyz(x^2+y^2+z^2)^{1/2}(1+1+1)^{1/2}=\sqrt3\,xyz. 
$$
Now we get the seemingly easier problem of maximizing $xyz$ under $x^2+y^2+z^2=1$. Here symmetry, common sense, or Lagrange multipliers show that the maximum is achieved when $x=y=z$. Then 
$$
xyz(x+y+z)\leq xyz(x^2+y^2+z^2)^{1/2}(1+1+1)^{1/2}=\sqrt3\,xyz\leq\frac{\sqrt3}{(\sqrt3)^3}=\frac13. 
$$
