<s>[INST] 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.
 [/INST] 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 $p=a+b+c$
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