The question is: $\forall x,y,z\in\Bbb R^+ \cup\{0\},\sum_{cyc}x=32$, find the maximum value of $\sum_{cyc}x^3y$.
This question was introduced in a "elementary" book talkng about inequalities.
The reason I put quotation marks is because this book contains mostly elementary knowledge,and solve not-so-elementary questions.
The method the book provides is to evaluate $27(\sum_{cyc} x)^4-256\sum_{cyc} x^3y$ and put it into form like $Ax^2+By^2$,so the upper bound is $110592$,it is equal iff $\{x,y,z\}=\{24,8,0\}$.(and you can guess how long that thing is)
This is clearly a method that I cannot think through myself. So I was wandering if someone can share a more elementary proof,just using normal inequalities(AM-GM,Cauchy,rearrangement,Chebyshev's sum,etc.).
Let $x\geq y\geq z$.
Thus, by AM-GM $$x^3y+y^3z+z^3x\leq(x+z)^3y=27\left(\frac{x+z}{3}\right)^3y\leq$$ $$\leq27\left(\frac{3\cdot\frac{x+z}{3}+y}{4}\right)^4=110592.$$ The equality occurs for $x=24$, $y=8$ and $z=0$, which says that in this case we got a maximal value.
The second case is similar.