If $abc=1$ and $a,b,c$ are positive real numbers, prove that $${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1\,.$$
The whole problem is in the title. If you wanna hear what I've tried, well, I've tried multiplying both sides by 3 and then using the homogenic mean. $${3 \over a+b+1} \le \sqrt[3]{{1\over ab}} = \sqrt[3]{c}$$ By adding the inequalities I get $$ {3 \over a+b+1} + {3 \over b+c+1} + {3 \over c+a+1} \le \sqrt[3]a + \sqrt[3]b + \sqrt[3]c$$ And then if I proof that that is less or equal to 3, then I've solved the problem. But the thing is, it's not less or equal to 3 (obviously, because you can think of a situation like $a=354$, $b={1\over 354}$ and $c=1$. Then the sum is a lot bigger than 3).
So everything that I try doesn't work. I'd like to get some ideas. Thanks.
Best Answer
let $$a=x^3,b=y^3,c=z^3\Longrightarrow xyz=1$$ since $$y^3+z^3\ge y^2z+yz^2$$ so $$\dfrac{1}{1+b+c}=\dfrac{xyz}{xyz+y^3+z^3}\le\dfrac{xyz}{xyz+y^2z+yz^2}=\dfrac{x}{x+y+z}$$ so $$\sum_{cyc}\dfrac{1}{1+b+c}\le\sum_{cyc}\dfrac{x}{x+y+z}=1$$