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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.

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  • $\begingroup$ 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? $\endgroup$
    – Calvin Lin
    Nov 11, 2021 at 21:02
  • $\begingroup$ $(x+1/x)(y+1/y) \geq (x+1/y)(y+1/x)$ 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. $\endgroup$
    – BodyDouble
    Nov 11, 2021 at 21:10
  • $\begingroup$ Great, do the same for 3 variables, and then AM-GM your way out of it. $\endgroup$
    – Calvin Lin
    Nov 11, 2021 at 21:11
  • $\begingroup$ 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? $\endgroup$
    – BodyDouble
    Nov 11, 2021 at 21:24
  • $\begingroup$ @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. $\endgroup$
    – BodyDouble
    Nov 11, 2021 at 21:43

2 Answers 2

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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.

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  • $\begingroup$ @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. $\endgroup$ Nov 11, 2021 at 21:03
  • $\begingroup$ 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. $\endgroup$
    – Calvin Lin
    Nov 11, 2021 at 21:10
  • $\begingroup$ Thank you, that makes sense. The rearrangement inequality was new to me. $\endgroup$
    – BodyDouble
    Nov 11, 2021 at 21:21
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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.

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