Prove that if $a+b+c=1$ then $\frac{ab}{c+1}+\frac{ac}{b+1}+\frac{bc}{a+1} \le \frac{1}4$ $a,b,c$ are positive real numbers

Question

I simplified the inequality , but I didn't get anything that may seem useful.

I got

$$^\sum_{cyc} 4a^2b+8ab+4ab^2\le abc+ab+ac+ab+a+b+c+1$$

Answer 1

For positive variables the reversed inequality is true.

Indeed, by C-S we obtain: $$\sum_{cyc}\frac{ab}{1+c}=\sum_{cyc}\frac{ab}{2c+a+b}\leq\frac{1}{4}\sum_{cyc}ab\left(\frac{1^2}{c+a}+\frac{1^2}{c+b}\right)=$$ $$=\frac{1}{4}\sum_{cyc}\left(\frac{ab}{c+a}+\frac{bc}{a+c}\right)=\frac{1}{4}\sum_{cyc}b=\frac{1}{4}.$$