Long Inequality problem for $a, b, c $ positive real numbers $$ \left( a+\frac{1}{b} -1\right) \left( b+\frac{1}{c} - 1\right) +\left( b+\frac{1}{c} -1\right) \left( c+\frac{1}{a} -1\right) +\left( c+\frac{1}{a} -1\right) \left( a+\frac{1}{b} -1\right) \geq 3$$
How we can prove the inequality   above.  Actually it take long time to prove it but I couldn't complete.  How we prove it? . Thanks for help
Maybe this could help:

If you denote $x=a+\frac{1}{b}&gt;0, \, y=b+\frac{1}{c}&gt;0, \, z=c+\frac{1}{a}&gt;0$
you will have $x+y+z=a+1/a +b+1/b+c+1/c\ge 2+2+2=6$ and the equality is achieved iff $a=b=c=1$

You get $(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)\ge 3$ which is equivalent to 
$$xy+yz+zx-2(x+y+z)\ge 0$$
So you have to prove the last one, having in mind that $x,y,z&gt;0$ and $x+y+z\ge 6$
Hint:

$a,b,c$ are positive real numbers.

So,$a+\frac 1b&gt;0$

So,$a+\frac 1b-1&gt;-1$.

Similarly all the values are  greater than $-1$.