Proving $\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1$

Question

$A,B,C$ 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

Answer 1

1. First, prove that if one big parenthesis is negative then the other two are positive, and the inequality is satisfied in this case.
2. 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.
3. obtain similar inequalities for $$\left(B-1-\frac{1}{C}\right)\left(C-1-\frac{1}{A}\right) $$ and $$\left(A-1-\frac{1}{B}\right) \left(C-1-\frac{1}{A}\right) $$
4. take the product of the resulting inequalities and you are done.

Answer 2

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 $(2)$?

Answer 3

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.