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.).

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

The second case is similar.

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  • $\begingroup$ Thank you so much. $\endgroup$ – StAKmod Feb 7 '19 at 20:46
  • $\begingroup$ @StAKmod You are welcome! $\endgroup$ – Michael Rozenberg Feb 7 '19 at 20:47
  • $\begingroup$ But how can I come up with this without knowing the answer?Do I just guess the answer and approach it? $\endgroup$ – StAKmod Feb 7 '19 at 20:48
  • $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Feb 7 '19 at 20:53
  • $\begingroup$ Thank you.I will look up to it and try to figure something out. $\endgroup$ – StAKmod Feb 7 '19 at 20:54

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