# How prove this inequality $x^3y+y^3z+z^3x\ge xyz(x+y+z)$

Let $x>0$, $y>0$ and $z>0$. Show that $$x^3y+y^3z+z^3x\ge xyz(x+y+z).$$

I known we can't WLOG: $x\ge y\ge z$, if this, I can use rearrangement inequality, But other I can't it. Thanks?

• Suppose for the time being $x\ge y\ge z$. Now observe that, \begin{align}(x^3y-x^2yz)+(y^3z-xy^2z)+(z^3x-xyz^2)&=x^2(xy-yz)+y^2(yz-zx)+z^2(zx-xy)\\&=(x^2-z^2)(xy-yz)+(y^2-z^2)(yz-zx)\\&=y(x^2-z^2)(x-z)+z(y^2-z^2)(y-z)\\&\ge0\end{align}To justify the last step we use the assumption that $x\ge y\ge z$ and the hypothesis that $x,y,z>0$. The other cases may be dealt in a similar manner. – user 170039 Aug 28 '17 at 15:42

## 2 Answers

You can not assume $x\geq y\geq z$ because our inequality is cyclic and not symmetric,

but we can say that $(x^2,y^2,z^2)$ and $\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)$ are opposite ordered.

Thus, by Rearrangement $$x^3y+y^3z+z^3x=xyz\left(\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\right)=$$ $$=xyz\left(x^2\cdot\frac{1}{z}+y^2\cdot\frac{1}{x}+z^2\cdot\frac{1}{y}\right)\geq$$ $$\geq xyz\left(\frac{x^2}{x}+\frac{y^2}{y}+\frac{z^2}{z}\right)=xyz(x+y+z).$$

C-S also works because $$\sum_{cyc}\frac{x^2}{z}\geq\frac{(x+y+z)^2}{x+y+z}=x+y+z,$$ but you wanted Rearrangement.

We can prove this inequality also by $uvw$, AM-GM, BW and more.

• so How to use Rearrangement? – wightahtl Aug 28 '17 at 14:20
• care to explain the first inequality? – windircurse Aug 28 '17 at 14:22
• I have use C-S by solve it,because $\sum \dfrac{z^2}{x}\sum x\ge (\sum x)^2$ – wightahtl Aug 28 '17 at 14:26
• @wightahtl windircurse I added something. I hope now it's clear. – Michael Rozenberg Aug 28 '17 at 14:27
• I believe @user170039 was trying to ask why you cannot assume $x\ge y\ge z$, though the sum of your responses have only come of the form that you don't need them, which doesn't answer the question at all. – Simply Beautiful Art Aug 28 '17 at 20:40

I don't know what is meant by rearrangement, so I'll offer another solution, fairly basic, but it works. $$x^3y+y^3z+z^3x \geq xyz(x+y+z)$$ $$xy(x^2)+yz(y^2)+zx(z^2) - xyz(x+y+z) \geq 0$$ $$xy(x^2-2xz+z^2)+yz(y^2-2yx+x^2)+zx(z^2-2yz+y^2) \geq 0$$ $$xy(x-z)^2 + yz(y-x)^2+zx(z-y)^2 \geq 0$$ The last line consists of all positive values, and thus we can work backwords and show this inequality to be true.

• – user 170039 Aug 28 '17 at 14:29
• Thanks! I will go learn about that now. – Isaac Browne Aug 28 '17 at 14:29
• @Isaac Browne Your proof it's just AM-GM, which I meant: $\sum\limits_{cyc}(x^3y+z^2xy)\geq\sum\limits_{cyc}\left(2\sqrt{x^3y\cdot z^2xy}\right)=2\sum\limits_{cyc}x^2yz$. – Michael Rozenberg Aug 28 '17 at 14:40
• @MichaelRozenberg Thanks for that, $(x-y)^2 \geq 0$ is indeed equivalent to AM-GM. I just like to think of this as more fundamental as we see all the positive values on one side being greater than 0. – Isaac Browne Aug 28 '17 at 14:46