Let $a,b,c$ be positive numbers such that $abc=1$. Prove that $\frac{a-1}{b}+\frac{b-1}{c}+\frac{c-1}{a} \geq 0$

Question

Question

Let $a,b,c$ be positive numbers such that $abc=1$. Prove that $$\frac{a-1}{b}+\frac{b-1}{c}+\frac{c-1}{a} \geq 0$$

My try

I have simplified this to the equivalent inequality $$ab^2+bc^2+ca^2 \geq ab+bc+ca$$

Now, I have tried AM-GM, but it does not work.

Any hint will be greatly appreciated.

Answer 1

By AM-GM : $ab^2+2bc^2\ge 3bc, ...$

Answer 2

Use Hölder and AMGM to prove that $RHS \leq (ab)^{3/2}+(bc)^{3/2}+(ca)^{3/2}=IS$.

So we want to prove $IS \leq LHS$.

$IS=\sqrt{ab}\sqrt{ab^2}+\ldots \leq \left(IS * LHS\right)^{1/2}$ by Cauchy-Schwarz.

Answer 3

Another approach: Let $$ X=\sum_{\text{cyc}}\frac{a}{b},\quad Y=\sum_{\text{cyc}}ab,\quad Z=\sum_{\text{cyc}}a. $$ Then we have $$ X+Y=\sum_{\text{cyc}}a\left(b+\frac{1}{b}\right)\ge 2\sum_{\text{cyc}}a=2Z, $$ and $$ XZ=\sum_{\text{cyc}}\frac{a}{b}\sum_{\text{cyc}}a=\sum_{\text{cyc}}a^2c\sum_{\text{cyc}}c\ge \left(\sum_{\text{cyc}}ac\right)^2=Y^2 $$ by AM-GM and Cauchy-Schwarz. Thus, $$ X(X+Y)\ge 2XZ\ge 2Y^2, $$and $$ (X+2Y)(X-Y)\ge 0. $$ This gives $$ \sum_{\text{cyc}}\frac{a}{b}=X\ge Y=\sum_{\text{cyc}}\frac{1}{b}. $$