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I am only being introduced to Olympiad math, and one particular aspect of it caught my attention in the book "Secrets in Inequalities".

It's $$\text{Let $x, y, z$ be positive real numbers. Prove that}$$ $$\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\ge 2+\frac{2(x+y+z)}{\sqrt[3]{xyz}}$$

Then the author claims that the problem follows this inequality

$$ \space \frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge \frac{x+y+z}{\sqrt[3]{xyz}} \qquad (1)$$

We can certainly rewrite the initial equation as $$\frac{x}{y}+\frac{x}{z}+\frac{y}{z} + \frac{y}{x}+\frac{z}{x}+\frac{z}{y} \ge \frac{2(x+y+z)}{\sqrt[3]{xyz}}$$

So, besides proving $(1)$, shouldn't we also prove $(2)$ $$ \space \frac{x}{z}+\frac{y}{x}+\frac{z}{y}\ge \frac{x+y+z}{\sqrt[3]{xyz}} \qquad (2)$$

or are they equivalent, and why?

Me myself tried to prove their equivalence using Jensen's inequality, assuming $f(n) = \frac{1}{n}$, like this:

if $(1)$ is true, then,

$$2) \space \frac{f\left(\frac{x}{y}\right) + f\left(\frac{y}{z}\right) + f\left(\frac{z}{x}\right)}{3} \ge f\left(\frac{\frac{x}{y}+\frac{y}{z}+\frac{z}{x}}{3} \right)$$

Unfortunately, it got to nothing

So, the chief question is why (1) and (2) are equivalent?

Michael Rozenberg
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Nurasik12
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    Given (1) is true for all positive $x, y, z$, it should hold true when replacing $(x, y, z)$ with $(x, z, y)$. The reverse is true similarly as well - so (1) and (2) are equivalent, these labels are swappable. – Macavity Jun 09 '25 at 11:45

3 Answers3

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From the inequality $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{x+y+z}{\sqrt[3]{xyz}}$$ You swap $y$ with $z$ to get $$\frac{x}{z}+\frac{z}{y}+\frac{y}{x}\ge\frac{x+z+y}{\sqrt[3]{xzy}}$$ $$\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\ge\frac{x+y+z}{\sqrt[3]{xyz}}$$ So $(1)$ and $(2)$ are equivalent.

Thinh Dinh
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  • Why can we swap characters and say that it is without loss of generality – Nurasik12 Jun 09 '25 at 10:12
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    Because the name of the variable is irrelevant to the inequality. We can even exchange $x$, $y$, $z$ into $a$, $b$, $c$. @Nurasik12 – youthdoo Jun 10 '25 at 05:18
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Another way.

We need to prove that: $$\prod_{cyc}(x+y)\geq 2xyz + 2(x+y+z) \sqrt[3]{x^2y^2z^2}$$ or $$\sum_{cyc}(x^2y+x^2z)\geq2\sum_{cyc}x^{\frac{5}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}},$$ which is true by Muirhead because $$(2,1,0)\succ\left(\frac{5}{3},\frac{2}{3},\frac{2}{3}\right).$$

Nurasik12
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Michael Rozenberg
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Expanding the LHS we have $$ \left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) = 2 + \left(\frac{x}{y}+\frac{x}{z}\right) + \left(\frac{y}{z}+\frac{y}{x}\right) + \left(\frac{z}{x}+\frac{z}{y}\right) $$ By am-gm, this becomes $$ \ge 2 + 2\frac{x}{\sqrt{yz}} + 2\frac{y}{\sqrt{zx}} + 2\frac{z}{\sqrt{xy}} $$ Therefore, we need to show $$ \frac{x}{\sqrt{yz}} + \frac{y}{\sqrt{zx}} + \frac{z}{\sqrt{xy}} \ge \frac{x+y+z}{\sqrt[3]{xyz}} $$

Multiplying by $\sqrt{xyz}$ gives $$ x^{3/2} + y^{3/2} + z^{3/2} \ge (x+y+z)(xyz)^{1/6} $$ Using power mean and am-gm

\begin{align} \frac{x^{3/2}+y^{3/2}+z^{3/2}}{3} &\ge \left(\frac{x+y+z}{3}\right)^{3/2} \\ &= \left(\frac{x+y+z}{3}\right) \sqrt{\frac{x+y+z}{3}} \\ &\ge \left(\frac{x+y+z}{3}\right) \sqrt{\sqrt[3]{xyz}} \\ &= \frac{x+y+z}{3} (xyz)^{1/6} \end{align}

Finish by multiplying by $3$, giving desired result.

Bumblebee
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