Let $a, b, c$ be positive real numbers such that $abc=1.$ Prove that $(a-1+1/b)(b-1+1/c)(c-1+1/a)\le1$

Question

Let $a, b, c$ be positive real numbers such that $abc=1.$ Prove that:

$$\left(a-1+\dfrac1b\right)\left(b-1+\dfrac1c\right)\left(c-1+\dfrac1a\right)\le1$$

or equivalently:

$$(ab-b+1)(bc-c+1)(ca-a+1)\le1$$

What I have tried:

Computing $\left(a-1+\dfrac1b\right)$ using $abc=1$ and similarly computing others and then multiplying them. But it didn't help.

Any help will be appreciated.

Note that you can keep multiplying by $abc$ to make the minus sign stand in front of any term you want. — Arthur, Sep 03 '16 at 07:08
I don't know whether this makes the solution any easier, but you can rewrite any positive real number $a$ as $e^\alpha$ for some real number $\alpha$. So you can put $a\equiv e^\alpha$, $b\equiv e^\beta$, $c\equiv e^\gamma$. Then, for example, $1/a = e^{-\alpha}$. — user326210, Sep 03 '16 at 07:15
You can use AM>=GM and a+1/a>=2 to get the result but it does not use abc=1. — sn24, Sep 03 '16 at 07:19
in the case when the condition is $$abc=1$$ we set $$a=x/y,b=y/z,c=z/x$$ — Dr. Sonnhard Graubner, Sep 03 '16 at 07:19

score 4 · Accepted Answer · edited Sep 03 '16 at 23:33

Note that at most one of the three factors is negative. For example if $$a-1+\frac1b<0\quad\text{and}\quad b-1+\frac1c<0,$$ then the first inequality implies $-1+\frac1b<0$ while the second implies $b-1<0$, an impossibility.

If exactly one factor is negative, then the product is negative and we are done. So assume all factors are positive. We have $$\left(a-1+\frac1b\right)\left(b-1+\frac1c\right)=ab-a+\frac ac-b+1-\frac1c+1-\frac1b+\frac1{bc}.$$ Using $ab=\frac1c$ and $\frac1{bc}=a$ this simplifies to $$\left(a-1+\frac1b\right)\left(b-1+\frac1c\right)=\frac ac-b-\frac1b+2\le \frac ac$$ since $b+\frac1b\geq 2$. Analogously we obtain $$\left(b-1+\frac1c\right)\left(c-1+\frac1a\right)\le \frac ba$$ and $$\left(c-1+\frac1a\right)\left(a-1+\frac1b\right)\le \frac cb.$$ The result follows by multiplying the three inequalities above. Equality holds if and only if $a=b=c=1$.

Let $a, b, c$ be positive real numbers such that $abc=1.$ Prove that $(a-1+1/b)(b-1+1/c)(c-1+1/a)\le1$

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