Inequality $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ Let $a,b,c\in\mathbb{R}^*_+$, $abc=1$.
How can i show that $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ ?
I got $\left(ab-b+1\right)\left(bc-c+1\right)\left(ca-a+1\right)\leq abc=1$, but i can't go any further ...
If i expand it : 
$2-ab-\frac{b}a-c-ac-\frac{a}c-b+2a+\frac{2}a-bc-\frac{c}b-a+2b+\frac{2}b+2c+\frac{2}c-4$
Which gives using $c=\frac{1}{ab}$ : 
$5ab-\frac{b}a-\frac{1}{ab}-a^2b+\frac{1}a+b+\frac{1}b-2$

Make the substitution $a=x/y$, $b=y/z$, $c=z/x$ where $x$, $y$ and $z$ are positive reals. The inequality becomes
$$
\left(\frac xy - 1 + \frac zy\right) \left( \frac yz - 1 + \frac xz \right) \left( \frac zx - 1 + \frac yx \right) \stackrel{?}{\leq} 1
$$
or
$$ (x-y+z)(x+y-z)(-x+y+z) \stackrel{?}{\leq} xyz. $$
At most one of the factors on the left hand side can be nonpositive and in such a case the inequality is obvious. If all factors are positive, the result follows by multiplying
$$ (x-y+z)(x+y-z) \leq \left(\frac{(x-y+z)+(x+y-z)}{2}\right)^2 = x^2 $$
and similar inequalities. The last inequality follows by AM-GM on $x-y+z$ and $x+y-z$.
