I have the following problem :
Let $a,b,c>0$ and $a\geq b \geq c $ then we have : $$\frac{1}{(a+b+c)}(\sum_{cyc}\frac{ab}{7a+b})\leq \frac{1}{8}$$
My proof :
We begin with two lemma :
First lemma : Let $a,b,c>0$ and $a\geq b \geq c $ then we have :$$(\sum_{cyc}\frac{a}{7a+b})\leq \frac{3}{8}$$
I didn’t prove it because it's easy .So the second lemma .
Second lemma : Let $a,b,c>0$ and $a\geq b \geq c $ then we have : $$\sum_{cyc}\frac{a^2}{7a+b}\geq \sum_{cyc}\frac{ac}{7a+b}\geq\sum_{cyc}\frac{ab}{7a+b}$$
It's an easy consequence of the rearrangement inequality .
Now the proof :
We have with the original condition and the first lemma :
$$\frac{(a+b+c)}{(a+b+c)}(\sum_{cyc}\frac{a}{7a+b})\leq \frac{3}{8}$$
Or :
$$\frac{1}{(a+b+c)}(\sum_{cyc}\frac{a^2}{7a+b}+\sum_{cyc}\frac{ac}{7a+b}+\sum_{cyc}\frac{ab}{7a+b})\leq \frac{3}{8}$$
With the second lemma we have : $$\frac{3}{(a+b+c)}(\sum_{cyc}\frac{ab}{7a+b})\leq \sum_{cyc}\frac{a^2}{7a+b}+\sum_{cyc}\frac{ac}{7a+b}+\sum_{cyc}\frac{ab}{7a+b}\leq \frac{3}{8}$$
Done !
My question is: Do you have an alternative proof ?
Thanks a lot for your time.
By C-S we obtain: $$\sum_{cyc}\frac{ab}{7a+b}=\sum_{cyc}\frac{1}{\frac{7}{b}+\frac{1}{a}}=\sum_{cyc}\frac{1}{\frac{49}{7b}+\frac{1}{a}}\leq\sum_{cyc}\frac{1}{\frac{64}{7b+a}}=\frac{a+b+c}{8}.$$