Proving $\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1$ $A,B,C$ are positive reals with product 1. Prove that $$\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1$$
How can I prove this inequality. I just need a hint to get me started. Thanks
 A: *

*First, prove that if one big parenthesis is negative then the other two are positive, and the inequality is satisfied in this case.

*Next, suppose that the three big parenthesis are positive.  Prove that 
$$\left(A-1-\frac{1}{B}\right) \left(B-1-\frac{1}{C}\right)\leq \frac{A}{C}$$
for this you do the product and you use the hint of abiessu inthe comment above.

*obtain similar inequalities for
$$\left(B-1-\frac{1}{C}\right)\left(C-1-\frac{1}{A}\right)  $$
and
$$\left(A-1-\frac{1}{B}\right) \left(C-1-\frac{1}{A}\right) $$

*take the product of the resulting inequalities and you are done.

A: Given that (w.l.o.g.) $0\lt A\le B\le C, A\cdot B\cdot C=1,$ we wish to show
$$\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1\tag 1$$
First, noting that $A-1+\dfrac 1B=AC+A-1$ and also $AC+A-1=(A-1)(C+1)+C,$ we transform $(1)$ as follows:
$$(AC+A-1)(AB+B-1)(BC+C-1)\\
=(A-1)(C+1)(B-1)(A+1)(C-1)(B+1)+C(AB+B-1)(BC+C-1)+A(A-1)(C+1)(BC+C-1)+B(A-1)(C+1)(A+1)(B-1)\\
=(A^2-1)(B^2-1)(C^2-1)+(1+BC-C)(BC+C-1)+(A-1)(C+1)(1+AC-A)+(1+BA-B-BC)(A+1)(B-1)$$
$$=(A^2-1)(B^2-1)(C^2-1)\\+B^2C^2-(1-C)^2+A^2C^2-(1-A)^2-C(1+AC-A)+\left(1-\frac 1A\right)(C+1)(A+1)(B-1)\tag 2$$
Can you take it forward from $(2)$?
A: Another way, substitute $A = \dfrac u v, B = \dfrac v w, C = \dfrac w u$ to get the equivalent inequality
$$\sum u^2v+\sum uv^2-\sum u^3\leqslant  3uvw$$
where the sums are cyclic.
But this is exactly Schur's inequality of degree $3$, hence proved. 
