 # Given 1+a+b+c=2abc and positivity of real numbers a,b,c, we are asked to prove that (ab)/(1+a+b)+(bc)/(1+b+c)+(ac)/(1+a+c)>=3/2 If d=a+b+c I got as far as to simplify the inequality into 1/a+1/b+1/c+1/(d−a)+1/(d−b)+1/(d−c)>=3 From a=(1+b+c)/(2bc−1) I also can prove that ab+ac+bc>=3/2 But cannot manage to get to the desired result. Massatfy 2022-08-16 Answered
Prove $\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ac}{1+a+c}\ge \frac{3}{2}$ for $a$, $b$, $c$ positive and $1+a+b+c=2abc$
If $d=a+b+c$ I got as far as to simplify the inequality into
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d-a}+\frac{1}{d-b}+\frac{1}{d-c}\ge 3$
From $a=\frac{1+b+c}{2bc-1}$ I also can prove that
$ab+ac+bc\ge \frac{3}{2}$
But cannot manage to get to the desired result.
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We need to prove next equivalent inequality
$\sum _{cyc}\left(\frac{ab}{1+a+b}+1\right)\ge \frac{9}{2}$
or
$\sum _{cyc}\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(1+a+b\right)\left(1+c\right)}\ge \frac{9}{2}$
Then by Cauchy-Schwarz inequality
$\sum _{cyc}\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(1+a+b\right)\left(1+c\right)}\ge \frac{9\prod _{cyc}\left(a+1\right)}{\sum _{cyc}\left(1+a+b\right)\left(c+1\right)}=$
$=\frac{9\prod _{cyc}\left(a+1\right)}{\sum _{cyc}\left(a+2ab+1+2a\right)}=\frac{9}{2}$
Now, we have done!