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Given $1+a+b+c = 2abc$ and positivity of real numbers $a,b,c$, we are asked to prove that $$\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ac}{1+a+c}\geq \frac{3}{2}$$

If $d=a+b+c$ I got as far as to simplify the inequality into $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d-a}+\dfrac{1}{d-b}+\dfrac{1}{d-c}\geq 3$$ From $a=\frac{1+b+c}{2bc-1}$ I also can prove that $$ab+ac+bc\geq \frac{3}{2}$$ But cannot manage to get to the desired result.

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    What is the source of this problem? – 2017-01-07
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    But even plugging in $a=b=c=1$ gives 1 on the left-hand side of the equation. – 2017-01-07
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    you can't put $a=b=c=1$. as in that case you break the condition. – 2017-01-07
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    @kolobokish. Then the condition should go into the title just as the other condition is. – 2017-01-07
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    @wilkersmon maybe you can try using the following $1 = 2abc - a -b-c$. I mean you can use it for $3$ and $2$. – 2017-01-07
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    @JimBaldwin Maybe.) – 2017-01-07

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We need to prove next equivalent inequality $$\sum\limits_{cyc}\left(\frac{ab}{1+a+b}+1\right)\geq\frac{9}{2}$$ or $$\sum\limits_{cyc}\frac{(a+1)(b+1)(c+1)}{(1+a+b)(1+c)}\geq\frac{9}{2}$$ Then by Cauchy-Schwarz inequality $$\sum\limits_{cyc}\frac{(a+1)(b+1)(c+1)}{(1+a+b)(1+c)}\geq\frac{9\prod\limits_{cyc}(a+1)}{\sum\limits_{cyc}(1+a+b)(c+1)}=$$ $$=\frac{9\prod\limits_{cyc}(a+1)}{\sum\limits_{cyc}(a+2ab+1+2a)}=\frac{9}{2}$$ Now, we have done!

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    Could you be a bit more thorough? I understood the first two inequalities but got totally lost further on. – 2017-01-07
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    @wilkersmon In the third line I used Cauchy-Schwarz: for positives $x$, $y$ and $z$ we have $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq\frac{(1+1+1)^2}{x+y+z}=\frac{9}{x+y+z}$ In the last line we have $\sum\limits_{cyc}(a+2ab+1+2a)=3(a+b+c)+2(ab+ac+bc)+3=2abc+2(ab+ac+bc)+2(a+b+c)+2=2(a+1)(b+1)(c+1)$. – 2017-01-07
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    Where can I find the proof of inequality $\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \frac{a_3^2}{b_3} \geq \frac{(a_1 + a_2 + a_3)^2}{b_1 + b_2 + b_3}$ ? I don't exactly get how this follows from C-S – 2017-01-07
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    @wilkersmon Because $\left(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}\right)(b_1+b_2+b_3)=$$\left(\left(\frac{a_1}{\sqrt{b_1}}\right)^2+\left(\frac{a_2}{\sqrt{b_2}}\right)^2+\left(\frac{a_3}{\sqrt{b_3}}\right)^2\right)\left(\left(\sqrt{b_1}\right)^2+\left(\sqrt{b_2}\right)^2+\left(\sqrt{b_3}\right)^2\right)\geq(a_1+a_2+a_3)^2$ – 2017-01-07
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    It's called C-S in Engel Form. – 2017-01-07