Inequalities (AMGM & Cauchy Schwarz)

Question

Let $ x, y, z \geq 0 $ such that $x+y+z=1$. Find the maximum value of $$x (x+y)^2 (y+z)^3 (z+x)^4.$$

Hi recently I've been stuck on this problem for quite some time, and I would appreciate some help/hints on how to approach this inequality.

Also, would be good if the approaches were based on AM-GM or Cauchy Schwarz.

Thanks!

Answer 1

Yes! We can solve this problem by AM-GM.

Indeed, $$x(x+y)^2 (y+z)^3 (z+x)^4=\frac{2x(2x+2y)^2(2y+2z)^3(z+x)^4}{64}\leq$$ $$\leq\frac{1}{64}\left(\frac{2x+2(2x+2y)+3(2y+2z)+4(z+x)}{10}\right)^{10}=\frac{1}{64}.$$ The equality occurs for $x=\frac{1}{2}$,$y=0$ and $z=\frac{1}{2}$, which says that the answer is $\frac{1}{64}.$

Done!