I was trying to prove the following inequality with the help of the rearrangement inequality: $$x, y, z > 0, xyz = 1 => xy^2 + yz^2 + zx^2 \geq x+y+z.$$
The first way I came up with was to multiply the RHS by $(xyz)^{2/3}$ which equals to 1 so that the inequality is ‘homogenized’. We get an inequality $xy^2 + yz^2 + zx^2 \geq x^{5/3}y^{2/3}z^{2/3} + x^{2/3}y^{5/3}z^{2/3} + x^{2/3}y^{2/3}z^{5/3}$ which is true due to the Muirhead’s inequality or the rearrangement inequality used for the two similarly ordered sequences $(xy^2)^{1/3}, (yz^2)^{1/3}, (zx^2)^{1/3}$ and $(xy^2)^{2/3}, (yz^2)^{2/3}, (zx^2)^{2/3}.$
This proof doesn’t feel ‘the rearrangement inequality’ enough for me so I came up with another one. It’s clear that either two of the numbers $x,y,z$ are $\leq 1$ and the third is $\geq 1$ or vice versa. The second case can be reduced to the first one by inverting the variables: let $X = 1/x, Y = 1/y, Z = 1/z.$ Numbers $X, Y, Z$ are positive and $XYZ = 1$ holds. Two of these numbers are $\leq 1$ so we can be sure that $XY^2 + YZ^2 + ZX^2 \geq X+Y+Z$. Then we can use the fact that $X+Y+Z \geq XY+YZ+ZX$ (this inequality is equivalent to $(X-1)(Y-1)(Z-1) \geq 0$ which is true). So we get: $XY^2 + YZ^2 + ZX^2 \geq XY+YZ+ZX.$ Going back to the original variables we deduce $xy^2 + yz^2 + zx^2 \geq x+y+z.$
Now, let’s prove the first case. The two numbers (WLOG $x$ and $y$) are $\leq 1$. Assume $1 \geq x \geq y$. Then we have two similarly ordered sequences: $$1 \geq x \geq y^2,$$ $$yz^2 \geq xz \geq x.$$ By the rearrangement inequality we get that $1 \cdot yz^2 + x \cdot xz + y^2 \cdot x \geq 1 \cdot x + x \cdot yz^2 + y^2 \cdot xz$ which is equivalent to $xy^2 + yz^2 + zx^2 \geq x+y+z.$
Now, what if $1 \geq y \geq x$? The inequality in question isn’t fully symmetrical so we can’t do quite the same. I was trying to find two similarly ordered sequences that will prove the inequality in this case but failed so far. Maybe someone here would help me with this!
Thanks for reading!
Best Answer
Here's a different solution:
Set $x = \frac{a}{b}, y = \frac{b}{c}, z = \frac{c}{a}$.
The inequality then transforms into:
$$\frac{ab}{c^2} + \frac{bc}{a^2} + \frac{ca}{b^2} \geq \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$$
Which can be rewritten as $$\frac{ab}{c^2} + \frac{bc}{a^2} + \frac{ca}{b^2} \geq \frac{ab}{b^2} + \frac{bc}{c^2} + \frac{ca}{a^2}$$
And from here the two similar sorted sequences $\{ab,ac,bc\}$ and $\{\frac{1}{c^2},\frac{1}{b^2},\frac{1}{a^2}\}$ should do the trick.