Let $a,b,c>0$ such that $a+b+c=3.$ Find the maximum $$P=\frac{a^2+7ab+b^2}{a+b+ab}+\frac{b^2+7bc+c^2}{b+c+cb}+\frac{c^2+7ac+a^2}{a+c+ac}.$$
I thought Max= 9 at $a=b=c=1.$
My attempt is using Cauchy-Schwarz inequality. It's
\begin{align*}
\sum_{\mathrm{cyc}}\frac{a^2+7ab+b^2}{a+b+ab}&=\sum_{\mathrm{cyc}}\frac{a^2+2ab+b^2+5ab}{a+b+ab}\\&=\sum_{\mathrm{cyc}}\frac{(a+b)^2}{a+b+ab}+5\sum_{\mathrm{cyc}}\frac{ab}{a+b+ab}.
\end{align*}
Now, $$9\sum_{\mathrm{cyc}}\frac{ab}{a+b+ab}\le \sum_{\mathrm{cyc}}ab\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}\right)=2(a+b+c)+3=9. $$
Thus, it's enough to prove $$\frac{(a+b)^2}{a+b+ab}+\frac{(c+b)^2}{c+b+cb}+\frac{(a+c)^2}{a+c+ac}\le 4.$$
But, it is reverse inequality since by Cauchy-Schwarz and
$ab+bc+ca\le 3$ $$\sum_{\mathrm{cyc}}\frac{(a+b)^2}{a+b+ab}\ge \frac{4(a+b+c)^2}{2(a+b+c)+ab+bc+ca}\ge 4$$
Is there others better idea? Hope you share it with us. Thank you.
Best Answer
$\sum_{cyc} \frac{a^2+7ab+b^2}{a+b+ab}=\sum_{cyc} \frac{(a+b)^2}{a+b+ab}+\frac{5ab}{a+b+ab}=(\sum_{cyc}a+b)+\sum_{cyc}( \frac{5ab}{a+b+ab}-\frac{(a+b)ab}{a+b+ab})$
=$6+\sum_{cyc} \frac{ab}{a+b+ab}{(5-(a+b))}$
We use chebyshevs inequality https://artofproblemsolving.com/wiki/index.php/Chebyshev%27s_Inequality
Understand the sequence $\frac{ab}{a+b+ab}$ is in oppsosite order to $5-(a+b)$
so thats $\leq 6+1/3 \sum_{cyc}(5-(a+b))\sum_{cyc}\frac{1}{\frac{1}{a}+\frac{1}{b}+1}$ $=6+3*\sum_{cyc} \frac{1}{\frac{1}{a}+\frac{1}{b}+1}$
by am gm [$\frac{1}{a}+\frac{1}{b}+1 \geq \frac{3}{(ab)^{1/3}}$] thats $\leq 6+\sum_{cyc} (ab)^{1/3}$
Now rest is just power mean inequality $\sum_{cyc} (ab)^{1/3} \leq 3 (\frac{ab+bc+ca}{3})^{1/3}$ and use $3(ab+bc+ca) \leq (a+b+c)^2$
answer would be $9$,value attained when $a=b=c=1$