# Find the maximum value $\frac{11a+4b}{4a^2-ab+2b^2}+\frac{11b+4c}{4b^2-bc+2c^2}+\frac{11c+4a}{4c^2-ca+2a^2}$

For the positive real numbers $a,b,c$ satisfy $ab+bc+ca=3abc$. Find the maximum value $$P=\frac{11a+4b}{4a^2-ab+2b^2}+\frac{11b+4c}{4b^2-bc+2c^2}+\frac{11c+4a}{4c^2-ca+2a^2}$$

i tried all methods such as: AM-GM; C-S,... but unsuccess. Help me give a hint

• Since both your constraint and the function you want to maximize are symmetric in the variables, I'd suspect the extrema must occur where the variables are the same, i.e. when $a=b=c=1$ and $P=9$. – Rus May Apr 5 '17 at 17:02
• If $a=b=c≒0, P→\infty$. – Takahiro Waki Apr 5 '17 at 19:05

If $a=b=c=1$ we get the value $9$.
Indeed, we need to prove that $$\sum_{cyc}\frac{11a+4b}{4a^2-ab+2b^2}\leq\frac{3(ab+ac+bc)}{abc}$$ or $$\sum_{cyc}\left(\frac{3}{a}-\frac{11a+4b}{4a^2-ab+2b^2}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)(a-6b)}{a(4a^2-ab+2b^2)}\geq0$$ or $$\sum_{cyc}\left(\frac{(a-b)(a-6b)}{a(4a^2-ab+2b^2)}+\frac{1}{b}-\frac{1}{a}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(a+b)}{ab(4a^2-ab+2b^2)}\geq0.$$ Done!