<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] Another way.
After full expanding we need to prove that:
$$\sum_{cyc}(a^2b+a^2c)\geq2(a+b+c)=2\sum_{cyc}a^{\frac{5}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}},$$ which is true by Muirhead because $$(2,1,0)\succ\left(\frac{5}{3},\frac{2}{3},\frac{2}{3}\right)$$ or by AM-GM:
$$\sum_{cyc}(a^2b+a^2c)\geq2\sum_{cyc}\sqrt{a^2b\cdot a^2c}=2(a+b+c).$$
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