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.

Answer 1

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

Answer 2

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

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