Problem. Let $a,b,c>0:abc=25.$ Find maximum $$M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$$
I've tried a new method $uvw$
When I set $a=b=5;c=1,$ calculate and I get $M=\dfrac{5}{66}.$ We will prove $$\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}\le \dfrac{5}{66}.$$
After full expanding, it remains to prove $f(u,v)\ge 0,$ where $a+b+c=3u; ab+bc+ca=3v^2; abc=w^3; u\ge v\ge 1.$
The task is consider fixed sign of $f'(u), f'(v)$ and we prove the inequality is true when two of equal variables.
I have just learnt about $uvw$ recently so I gave brief idea.
Am I in right approach? Please help me find troubles if my idea is wrong.
By the way, I also hope to see others proof. Thank you very much.
Best Answer
For $a=b=5$ and $c=1$ we obtain a value $\frac{5}{66}.$
We'll prove that it's a maximal value.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $$\sum_{cyc}\frac{1}{a+b+12c}=\sum_{cyc}\frac{1}{3u+11c}=\frac{\sum\limits_{cyc}(3u+11c)(3u+11b)}{\prod\limits_{cyc}(3u+11a)}=$$ $$=\frac{225u^2+363v^2}{324u^3+1089uv^2+1331w^3}$$ and we need to prove that $$\frac{225u^2+363v^2}{324u^3+1089uv^2+1331w^3}\leq \frac{5}{66}$$ or $f(v^2)\geq0,$ where $f$ is a linear function.
But the linear function gets a minimal value for an extremal value of $v^2$, which by $uvw$ happens for equality case of two variables.
Let $b=a$ and $c=\frac{25}{a^2}.$
Thus, it's enough to prove that: $$\frac{2}{13a+\frac{25}{a^2}}+\frac{1}{2a+\frac{300}{a^2}}\leq\frac{5}{66},$$ which gives $$(a-5)^2(64a^4+89a^3-735a^2+300a+750)\geq0,$$ which is true because by AM-GM: $$64a^4+89a^3-735a^2+300a+750\geq\left(2\sqrt{64\cdot750}+2\sqrt{89\cdot300}-735\right)a^2>$$ $$>\left(2\sqrt{64\cdot729}+2\sqrt{81\cdot289}-735\right)a^2=3a^2>0.$$