# How do I prove this inequality?: $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ where $a,b,c>0$ and $abc=1.$

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.

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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.

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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.

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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 $a=b=c=1$

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