If $abc=25,$ find maximum $M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$

Question

Problem. Let $a,b,c>0:abc=25.$ Find maximum $$M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$$

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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.

Answer 1

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: $$64a^4+89a^3-735a^2+300a+750\geq\left(2\sqrt{64\cdot750}+2\sqrt{89\cdot300}-735\right)a^2>$$ $$>\left(2\sqrt{64\cdot729}+2\sqrt{81\cdot289}-735\right)a^2=3a^2>0.$$

Answer 2

Calculus approach (Method of Lagrange's multiplier).

Let $M(a,b,c)=\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$ where $A=b+c+12a$, $B=a+c+12b$, $C=a+b+12c$ and let $g(a,b,c)=abc-25$. We Eliminating $\lambda$ in the system $\frac{\partial M}{\partial i}=\lambda\frac{\partial g}{\partial i}$, $i=a,b,c$. We obtain the cyclic system $$cyc: (13bc-ab-cA)A^2=(13ac-ab-bc)B^2$$ and by using $B^2-A^2=(B-A)(B+A)$, we have the system $$cyc:\\ 11(b-a)(2b+13c+13a)(13ac-ab-bc)=14c(b-a)A^2$$ Hence, $b=a$ leads to extrema. But, I still couldn't prove that all is such.

And $b=a$ leads to the equation $$221a^9-34650a^6+915625a^3-4687500=0$$ which has three positive real roots. The largest $a=5$ gives the maximum of the function.