4
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$x$,$y$,$z$ are non-negative numbers.
$x+y+z=3$

Find the maximum value of $~$ $x^{2}y+y^{2}z+z^{2}x$ $~$ without calculus.

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5
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in general

if $x,y,z\ge 0$,and such $x+y+z=3$, then we have $$x^ky+y^kz+z^kx\le\max{\{3,\dfrac{3^{k+1}k^k}{(k+1)^{k+1}}\}}$$

For $k=2$ I have nice methods

with out loss of let $x=\max{(x,y,z)}$

then we use Benoulli inequality,we have $$(1+\dfrac{z}{x})^2\ge 1+\dfrac{2z}{x}$$ so $$(x+z)^2y=x^2y(1+\frac zx)^2\ge x^2y(1+\dfrac{2z}{x})=x^2y+xyz+xyz\ge x^2y+y\cdot yz+xz^2$$ so $$x^2y+y^2z+z^2x\le (x+z)^2y=2^2\left(\dfrac{x+z}{2}\right)^2\cdot y\le 2^2\left(\dfrac{x+z+y}{3}\right)^3=4$$

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  • $\begingroup$ Thanks for the nice solution. $\endgroup$ – chloe_shi Sep 10 '13 at 10:21
0
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Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.

Hence, by Rearrangement and AM-GM we obtain: $$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$ $$=b(a^2+ac+c^2)\leq b(a^2+2ac+c^2=b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq$$ $$\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3=4.$$ The equality occurs for $z=0$, $x=2$ and $y=1$, which says that the answer is $4$.

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