Prove the inequality $xyz^3 \leq 27(\frac{x+y+z}{5})^5$. For $x,y,z>0$, prove the inequality $xyz^3 \leq 27(\frac{x+y+z}{5})^5$.
Any ideas? I am stuck.

Write $f(x,y,z) = 27 \left( \frac{x+y+z}{5} \right)^5 - x y z^3$. Then
$$\frac{\partial f}{\partial x}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - y z^3$$
$$\frac{\partial f}{\partial y}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - x z^3$$
$$\frac{\partial f}{\partial z}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - 3 x y z^2$$
At a minimum these all should equal $0$, so $yz^3 = xz^3 = 3x y z^2$, which gives us $x=y$ and $z=3x$. Plugging these in the inequality shows that the equality is attained at these points. One also has to check the boundary, i.e. when one or two or three of the variables is $0$, but this is trivial.
