Prove that $\frac1{a+b+1}+\frac1{b+c+1}+\frac1{c+a+1}\le1$

Question

If $abc=1$ then

$$\frac1{a+b+1}+\frac1{b+c+1}+\frac1{c+a+1}\le1$$

I have tried AM-GM and C-S and can't seem to find a solution. What is the best way to prove it?

Answer 1

Also we can use the following reasoning.

For positives $a$, $b$ and $c$ let $a=x^3$, $b=y^3$ and $c=z^3$.

Hence, $\sum\limits_{cyc}\frac{1}{a+b+1}=\sum\limits_{cyc}\frac{1}{x^3+y^3+xyz}\leq\sum\limits_{cyc}\frac{1}{x^2y+y^2x+xyz}=\sum\limits_{cyc}\frac{z}{x+y+z}=1$,

but I think the first way is better.

Answer 2

It's wrong, of course.

For positives $a$, $b$ and $c$ it's $\sum\limits_{cyc}(a^2b+a^2c-2a)\geq0$, which is Muirhead because $(2,1,0)\succ\left(\frac{5}{3},\frac{2}{3},\frac{2}{3}\right)$.