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Given three positive real numbers $x, y, z$ so that $x+ y+ z= 3$, prove that: $$x^{2}y+ y^{2}z+ z^{2}x\leq 4$$ I homogenized the inequality, so I set $x+ y+ z= 3$ to simplify the expression.

Now I'm stuck. I noticed an interesting equality case:

At $\left ( x, y, z \right )= \left ( 2, 1, 0 \right )$, equality holds. However, when $x= y= z= 1$, the expression evaluates to only $3$, so equality does not occur there.

To me, this suggested some kind of weighted AM-GM might apply, but I couldn't get it to work.

Thanks in advance for any help!

Dang Dang
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John Marty
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5 Answers5

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Assume to have a fixed $z\in(0,3]$, then maximize $f(x,y)=x^2 y + y^2 z + z^2 x$ under the constraints $x+y=3-z$ and $x,y>0$. This is the same as maximizing $$ g(x)=-x^3 + 3 x^2 + (3 z^2 - 6 z) x + (9 z + z^3) $$ under the only constraint $x\in(0,3-z)$. Since: $$ g'(x) = -3((x-1)^2-(z-1)^2)=3(z-x)(x+z-2)$$ the stationary points of $g(x)$ occur when $x=z$ or $x+z=2$. This gives that it is sufficient to prove the starting inequality when $(x,y,z)=(z,3-2z,z)$ with $z\in(0,3/2)$ and when $(x,y,z)=(2-z,1,z)$ with $z\in(0,2)$. In the first case, we have to prove: $$ z\in(0,3/2)\quad\Longrightarrow\quad 3z^3-9z^2+9z\leq 4$$ that is straightforward since $\frac{d}{dz}(3z^3-9z^2+9z)=3(z-1)^2\geq 0$ and the inequality holds for $z=\frac{3}{2}$. In the latter case, we have to prove: $$ z\in(0,2)\quad\Longrightarrow\quad z(z^2-3z+3)\geq 0$$ that is also straightforward.

Jack D'Aurizio
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4

Use the well-known inequality $$4\left ( x+ y+ z \right )^{3}\geq 27\left ( xy^{2}+ yz^{2}+ zx^{2}+ xyz \right )$$

Dang Dang
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math110
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3

It suffices to show that

$$4(\,x+ y+ z\,)^{\,3}- 27(\,x^{\,2} y+ y^{\,2}z+ z^{\,2}x+ xyz\,)\geqslant 0$$

Can let

$$F= 4(\,x+ y+ z\,)^{\,3}- 27(\,x^{\,2} y+ y^{\,2}z+ z^{\,2}x+ xyz\,)$$

$$\therefore F= (\,x+ 4\,z\,)(\,y+ z- 2\,x\,)^{\,2}+ 4\,y(\,y- z\,)^{\,2}- 11y(\,x- y\,)(\,x- z\,)\geqslant 0$$

$$\because\,x\equiv \text{mid}\{x,\,y,\,z\} \tag{assume}$$

NKellira
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this is not a solution to your problem, but may shed a little light on the fact you found puzzling

constrained optimization finds the only critical point is $(1,1,1)$. this is a minimum, so the maximum occurs on the boundary.

$$ \frac{\partial}{\partial x}\left(x^2y+y^2z+z^2x -\lambda(x+y+z) \right) =0 $$ gives: $$ 2xy + z^2 = \lambda $$ subtracting this from the two similar equations for $y,z$ gives $x=y=z$

now, setting one of the variables, say $y$ to zero, for the boundary, gives, by a similar procedure: $$ 2zx=\lambda = z^2 $$ one of whose two solutions gives the maximum you noticed at $(1,0,2)$.

David Holden
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Using Ji Chen's decomposition.

Assuming $x= \min\left\{ x, y, z \right\}$. $$4(\sum\limits_{cyc}x)^{3}- 27\sum\limits_{cyc}xy^{2}= \left ( x+ y- 2z \right )^{2}\left ( 4y+ z- 5x \right )+ 9x\sum\limits_{cyc}\left ( x^{2}- xy \right )\geq 0.$$

Dang Dang
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