Prove that, for all non-negative real numbers $x,y,z$ that satisfy $x+y+z=1, x^2y+y^2z+z^2x≤4/27$

Question

Prove that, for all non-negative real numbers $x, y, z$ that satisfy $x + y + z = 1$, $$x^2 y + y^2 z + z^2 x \leq \frac {4}{27} $$

I'm having trouble with this question. I suspect it may have a fairly simple proof using the AM-GM inequality and certain substitutions, however, I have been unable to complete such a proof.

See http://artofproblemsolving.com/community/c6h216p768 . (The topic includes various generalizations, which I find significantly more interesting than the problem itself.) — darij grinberg, Nov 05 '16 at 04:26

score 2 · Answer 1 · edited Jan 21 '17 at 11:02

2

Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.

Hence, by Rearrangement and AM-GM we obtain: $$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$ $$=b(a^2+ac+c^2)\leq b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}$$

edited Jan 21 '17 at 11:02

miracle173

11,359

answered Jan 21 '17 at 05:28

Michael Rozenberg

203,855

Why do you introduce new variables and not assume w.l.o.g $x\ge y \ge z$. That would be the same. 2. You can't assume w.l.o.g $x\ge y \ge z$ or use your a,b,c argument. There is no symmetry. 3. Why do you need $a \ge b \ge c$? I think this is not necessary.

miracle173

Jan 21 '17 at 07:25

@miracle173 1. This inequality is cyclic and not symmetric. I can not assume $x\geq y\geq z$. 2. I can assume $\max{x,y,z}=a$, $\min{x,y,z}=c$ and the last variable equal to $b$, which gives a permutation of $x$, $y$ and $z$ such that $a\geq b\geq c$ and we get possibility to use Rearrangement. – Michael Rozenberg Jan 21 '17 at 08:59

@miracle173 I am waiting for your reaction. – Michael Rozenberg Jan 21 '17 at 10:47

sorry, i missed that $x\cdot xy+y\cdot yz+z\cdot zx$ and $a\cdot ab+b\cdot ac+c\cdot bc$ are different. I now see the difference and you are right. – miracle173 Jan 21 '17 at 11:05

score -1 · Answer 2 · answered Nov 05 '16 at 05:06

WLOG, assume $x\geq y\geq z$. You wish to maximize the function: $f(x,y,z)=x^2y+y^2z+z^2x$ given that $x,y,z \in \Re^{+} \cup\{0\}$ and $x+y+z=1$

Now, note that(we're trying to start of by obtaining a trivial inequality based on our assumption: $$\begin{split}f(x+z,y,0)-f(x,y,z)= & (x+z)^2y-(x^2y+y^2z+z^2x)=\\ & z^2y + yz(x - y) + xz(y - z) \geq 0\end{split}$$ This means that we can assume value of $z$ to be $0$. Letting $z=0 (\implies x+y=1)$ in our original function and applying AM-M inequality with the terms $x,x,2y$: $$f(x,y,0)=x^2y=2\frac{x^2y}{2}\leq\left(\frac{x+x+2y}{3}\right)^3=\left(\frac{2(x+y)}{3}\right)^3=\frac{4}{27}$$

The inequality is NOT symmetric, it is cyclic, so you cannot always assume $x\ge y\ge z$. You have to also consider the possibility $x\ge z\ge y$ for instance, where your trivial inequality isn't so trivial anymore. — Macavity, Nov 05 '16 at 09:26

Prove that, for all non-negative real numbers $x,y,z$ that satisfy $x+y+z=1, x^2y+y^2z+z^2x≤4/27$

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