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

Solution 1:

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.

Solution 2:

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.