# Finding the upper bound of a 3-variable function.(inequality)

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.).

• What is $\sum_{cyc}$? – gt6989b Feb 7 '19 at 20:30
• @gt6989b $\sum_{cyc} x^3y=x^3y+y^3z+z^3x$ – StAKmod Feb 7 '19 at 20:36

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.
• @StAKmod It comes from an experience. But now it's very difficult to explain because you showed the answer. There is also known problem from Canadian Olimpiad: For non-negatives $x$, $y$ and $z$ such that $x+y+z=3$ prove that $x^2y+y^2z+z^2x+xyz\leq4.$ In this problem happen similar things. – Michael Rozenberg Feb 7 '19 at 20:53