Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been thinking over this problem for a couple of days, but I have no idea how to solve it in a simple way. I am interested if there is a way only using elementary methods to prove it. Using the software Mathematica confirmed this inequality is correct.

share|cite|improve this question
up vote 11 down vote accepted

Since $abc = 1$, any product term such as $ab$ or $\frac{1}{bc}$ can be rewritten as a singleton term (i.e., $ab = \frac{1}{c}$ and $\frac{1}{bc} =a$. Thus the inequality is equivalent to $$\left(a-1+\frac{1}{c}\right)\left(b-1+\frac{1}{a}\right)\left(c-1+\frac{1}{b}\right) \leq 1.$$

This inequality was Problem 2 on the 2000 International Mathematical Olympiad. Knowing that, solutions should be easy to find.

For example, here's a nice one due to Robin Chapman that can be found here.

We have $$\left(b - 1 + \frac{1}{a}\right) = b\left(1 - \frac{1}{b} + \frac{1}{ab}\right) = b\left(1 + c - \frac{1}{b}\right).$$ Hence, $$\left(c - 1 + \frac{1}{b}\right)\left(b - 1 + \frac{1}{a}\right) = b\left(c^2 - \left(1 - \frac{1}{b}\right)^2\right) \leq b c^2.$$ Thus $$\left(a-1+\frac{1}{c}\right)^2\left(b-1+\frac{1}{a}\right)^2\left(c-1+\frac{1}{b}\right)^2 \leq b c^2 a b^2 c a^2 = 1,$$ proving the inequality in the case that every factor on the left side of the inequality is positive.

Now, suppose one of the factors on the left-hand side is negative; i.e., $a - 1 + \frac{1}{c} < 0$. Then $a < 1$ and $c > 1$. Thus $b-1+\frac{1}{a} > 0$ and $c-1+\frac{1}{b} > 0$. So the left-hand side of the inequality is negative, and the inequality still holds.

share|cite|improve this answer
A beautiful proof! – Eastsun May 5 '11 at 6:42
Well, just one of the factors cannot be negative because then $abc \ne 1$. But two of them could, and then your remark at the end still works. – Glen Wheeler May 5 '11 at 7:37

A good start would be to show that the only extremum is at $a=b=c=1$. First, we put $c=1/ab$, and from now on we can ignore $c$. Second, we calculate the derivatives of $f = RHS-LHS$ with respect to $a,b$. We get some rational expressions, for which we can extract the nominators (the denominator turns out to be positive, if it helps). We can equate both to $0$, and using Groebner bases find that the only extremum is at $a=b=1$.

There are now several ways to complete the proof. For example, it may be possible to prove that unless $a,b$ are both in some interval $[l,h]$ then obviously $f > 0$. The function $f$ must then attain its minimum inside the interval, which must be the extremum we calculated.

share|cite|improve this answer

With $abc=1$, we can set :$a=\frac{y}{x}; b=\frac{z}{y}; c=\frac{x}{z}$. We have:

$\frac{x}{z}+\frac{y}{z}+\frac{z}{y}+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\le 3+\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}$

$\Leftrightarrow 3xyz+x^3+y^3+z^3\ge xy(x+y)+yz(y+z)+zx(z+x)$.

It's true by Schur's inequality. Equality occurs if only if $a=b=c=1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.