Inequality $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$

Solution 1:

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 $x+y-z$.