1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 1. 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. $\endgroup$ – miracle173 Jan 21 '17 at 7:25
  • $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Jan 21 '17 at 8:59
  • $\begingroup$ @miracle173 I am waiting for your reaction. $\endgroup$ – Michael Rozenberg Jan 21 '17 at 10:47
  • $\begingroup$ 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. $\endgroup$ – miracle173 Jan 21 '17 at 11:05
-1
$\begingroup$

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 $x,x,2y$: $$f(x,y,0)=x^2y=2\frac{x^2y}{2}\leq\left(\frac{x+x+2y}{3}\right)^3=\left(\frac{2(x+y)}{3}\right)^3=\frac{4}{27}$$

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Macavity Nov 5 '16 at 9:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.