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.
 A: 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
$$\sum_{cyc}\dfrac{1}{1+b+c}\le\sum_{cyc}\dfrac{x}{x+y+z}=1$$
A: 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
$$\frac{1}{1+x^3+y^3}+\frac{1}{1+y^3+z^3}+\frac{1}{1+z^3+x^3} \leq \frac{1}{1+xy(x+y)}+\frac{1}{1+xz(x+z)}+\frac{1}{1+yz(z+y)} \\ \leq \frac{1}{1+\frac{1}{z}(x+y)}+\frac{1}{1+\frac{1}{y}(x+z)}+\frac{1}{1+\frac{1}{x}(z+y)}=1$$
A: 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 $(c-1)^2(a+b) \ge 0$.
A: 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 $p=a+b+c$
A: 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:
$$\sum_{cyc}(a^2b+a^2c)\geq2\sum_{cyc}\sqrt{a^2b\cdot a^2c}=2(a+b+c).$$
A: Using the Cauchy-Schwarz inequality we have
$$\frac{1}{1+a+b}=\frac{c+2}{(c+1+1)(1+a+b)} \leqslant \frac{c+2}{(\sqrt{c}+\sqrt{a}+\sqrt{b})^2}.$$
Therefore
$$\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leqslant \frac{a+b+c+6}{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}.$$
But $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2 = a+b+c+2(\sqrt{ab}+\sqrt{ab}+\sqrt{ab}) \geqslant a+b+c+6.$$
So
$$\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leqslant 1.$$
