If $abc=1$ and $a,b,c$ are positive real numbers, prove that ${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1$.

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.

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 $$\sum_{cyc}\dfrac{1}{1+b+c}\le\sum_{cyc}\dfrac{x}{x+y+z}=1$$

This answer only assumes that $abc\geq 1$. Make the following substitution $$\sqrt[3]{a}=x,\sqrt[3]{b}=y,\sqrt[3]{c}=z$$ then we have $xyz\geq1$ and we have to prove the following inequality now

$$\frac{1}{1+x^3+y^3}+\frac{1}{1+y^3+z^3}+\frac{1}{1+z^3+x^3} \leq 1 $$

Clearly $$(x^3+y^3)=(x+y)(x^2-xy+y^2)\overset{\text{AM-GM}}{\geq}(x+y)xy$$

We have the following chain of inequalities

$$\frac{1}{1+x^3+y^3}+\frac{1}{1+y^3+z^3}+\frac{1}{1+z^3+x^3} \leq \frac{1}{1+xy(x+y)}+\frac{1}{1+xz(x+z)}+\frac{1}{1+yz(z+y)} \\ \leq \frac{1}{1+\frac{1}{z}(x+y)}+\frac{1}{1+\frac{1}{y}(x+z)}+\frac{1}{1+\frac{1}{x}(z+y)}=1$$

For my earlier comment: By expanding everything I mean, you can clear the denominator, write down everything in terms of symmetric polynomials, and try to use AM-GM to compare them.

On the other hand, there is also a one liner, similar to math110's solution:

$$\frac{1}{a+b+1} \leq \frac{2c+ab}{2(a+b+c)+ab+bc+ca}$$

After clearing the denominator, this is equivalent to $(c-1)^2(a+b) \ge 0$.

I have other nice Cauchy-Schwarz inequality solve it.

since $$\dfrac{1}{1+a+b}=1-\dfrac{a+b}{1+a+b}$$ so the original inequality can be written $$\sum_{cyc}\dfrac{a+b}{a+b+1}\ge2$$ use Cauchy-Schwarz inequaliy and the AM-GM inequality,we have $$\sum_{cyc}\dfrac{a+b}{a+b+1}\ge\dfrac{(\sum\sqrt{a+b})^2}{\sum(a+b+1)}=\dfrac{2p+2\sum\sqrt{(a+b)(a+c)}}{2p+3}\ge\dfrac{2p+2\sum(a+\sqrt{bc})}{2p+3}=\dfrac{4p+2\sum\sqrt{bc}}{2p+3}\ge 2$$ because use AM-GM inequality $$\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\ge 3\sqrt[3]{abc}=3$$ where $p=a+b+c$