# Proving an inequality with 3 variables [closed]

How would I go about proving the following? Let $$x,y,z$$ be positive real numbers, then $$\left(x + \frac{1}{x} \right)\left(y + \frac{1}{y} \right)\left(z + \frac{1}{z} \right) \geq \left(x + \frac{1}{y} \right)\left(y + \frac{1}{z} \right)\left(z + \frac{1}{x} \right).$$ Proving this for only 2 variables is simple, but with 3 I'm stumped. Expanding the expressions becomes messy and I see no clear way to proceed.

• Can you explain how you did it for 2 variables? Thereafter, how did you try to extend it to 3 variables and couldn't seem to push through? Nov 11, 2021 at 21:02
• $(x+1/x)(y+1/y) \geq (x+1/y)(y+1/x)$ becomes $xy + x/y + y/x + 1/(xy) \geq xy + 1/(xy) + 2$, which becomes $x^2 + y^2 \geq 2xy$, and finally $x^2 - 2xy + y^2 = (x-y)^2 \geq 0$. The problem when expanding to 3 variables is that now $z$ is involved in the second multiplicand term in the RHS. Nov 11, 2021 at 21:10
• Great, do the same for 3 variables, and then AM-GM your way out of it. Nov 11, 2021 at 21:11
• The AM-GM inequality is new to me. Looking at the Wikipedia article it looks like I would have to use some cubic root. Since i 'completed the square' in the case for 2 variables, do I have to complete the cube here? Nov 11, 2021 at 21:24
• @CalvinLin can you elaborate? It would be interesting to see another way to prove this, if it's possible. I tried to do the same for 3 variables as for 2, but I see no apparent to line the terms up in such a way to use AM-GM. Nov 11, 2021 at 21:43

Here's a sketch. Suppose that $$x\leqslant y\leqslant z$$. Use rearrangement inequality on the following pairs of increasing sequences:

1. $$(x,y,z)$$ and $$(\frac{1}{yz},\frac{1}{zx},\frac{1}{xy})$$

2. $$(xy,xz,yz)$$ and $$(\frac{1}{z},\frac{1}{y},\frac{1}{x})$$

We get the following respectively:

1.$$\;\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leqslant\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}$$

2.$$\;x+y+z \leqslant \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$$

From the above inequalities, we have:

$$x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leqslant \frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}+\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$$

Adding $$xyz+\frac{1}{xyz}$$ to both sides of the above inequality, we get the required inequality.

Now suppose that $$x\leqslant z\leqslant y$$. Use rearrangement inequality on the following pairs of increasing sequences:

1. $$(x,z,y)$$ and $$(\frac{1}{yz},\frac{1}{xy},\frac{1}{xz})$$

2. $$(xz,xy,yz)$$ and $$(\frac{1}{y},\frac{1}{z},\frac{1}{x})$$

We get the following respectively:

3.$$\;\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leqslant\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}$$

4.$$\;x+y+z \leqslant \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$$

From the above inequalities, we have:

$$x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leqslant \frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}+\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$$

Adding $$xyz+\frac{1}{xyz}$$ to both sides of the above inequality, we get the required inequality. Without loss of generality, it is sufficient to check for these chains.

• @Calvin Lin Thank you. I meant to suggest that the other case follows similarly, but should've mentioned it. Let me make the required edit. Nov 11, 2021 at 21:03
• As an aside, the 4 (and higher) variable case also holds. For those, we can't easily WLOG our way out, but the ideas expressed here could be extended to apply there. Nov 11, 2021 at 21:10
• Thank you, that makes sense. The rearrangement inequality was new to me. Nov 11, 2021 at 21:21

3 variables: Expanding, we want to show that

$$\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} + \frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy} \geq x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}.$$

This is true by taking sums over the following AM-GM inequalities:

• $$\frac{xy}{z} + \frac{yz}{x} \geq 2 \sqrt{ y^2} = 2 y$$ and cyclic versions.
• $$\frac{x}{yz} + \frac{y}{zx} \geq 2 \sqrt{ \frac{1}{ z^2} } = 2\frac{1}{z}$$ and cyclic versions.

Note: The result (and generalized result) follows directly from the statement of the Reverse Rearrangement Inequality. But given that OP hasn't heard of Rearrangement, I was going to just stick to AM-GM.