Max $P= \frac{2ab+3b^2}{(a+3b)^2}+\frac{2bc+3c^2}{(b+3c)^2}+\frac{2ca+a^2}{(c+3a)^2}$

Question

Let: $a,b,c>0$. Find the maximum value of: $$P= \frac{2ab+3b^2}{(a+3b)^2}+\frac{2bc+3c^2}{(b+3c)^2}+\frac{2ca+a^2}{(c+3a)^2}$$

Here are my try:

I tried to use tangent line trick, then I got:

$$\frac{2ab+3b^2}{(a+3b)^2} \leq \frac{1}{4}(a-b)+\frac{5}{16}$$

which is only true if $a \geq b$

I tried to use Cauchy - Schwarz but it seem impossible.

Answer 1

Let $\dfrac{a}{b}, \dfrac bc, \dfrac ca = x,y,z$ respectively, so you have $xyz=1, x,y, z >0.$ Then, prove: $$\dfrac{2x+3}{(x+3)^2}\leq\dfrac{11-x}{32}$$ and after summing up, you just need to show: $$x+y+z\geq 3$$ which is just AM-GM.

The comment below is right - this is only good for $x,y,z\leq 3.$ But if one of them, say $x$, is larger than $3,$ then: $$\dfrac{2x+3}{(x+3)^2}<\dfrac{9}{36}$$ and so: $$\sum_{x,y,z} \dfrac{2x+3}{(x+3)^2}< \dfrac{9}{36} + \dfrac 13 + \dfrac 13 = \dfrac{11}{12}<\dfrac{15}{16},$$ since $$\dfrac{2x+3}{(x+3)^2} = \dfrac{2x+3}{x^2+3(2x+3)}<\dfrac{1}{3}.$$

Answer 2

For $a=b=c=1$ we obtain a value $\frac{15}{16}.$

We'll prove that it's a maximal value.

Indeed, let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.

Thus, $xyz=1$ and we need to prove that: $$\sum_{cyc}\frac{2x+3}{(x+3)^2}\leq\frac{15}{16}$$ or $$\sum_{cyc}\left(\frac{5}{16}-\frac{2x+3}{(x+3)^2}-\frac{1}{32}\ln{x}\right)\geq0.$$

Now, let $f(x)=\frac{5}{16}-\frac{2x+3}{(x+3)^2}-\frac{1}{32}\ln{x}.$

Thus, $$f'(x)=\frac{(x-1)(27+54x-x^2)}{32x(x+3)^3},$$ which says that $x_{min}=1$, $x_{max}=27+\sqrt{758}$ and since $f\left(x_{max}\right)>0,$ there is an unique $x_1>x_{max},$ for which $f(x_1)=0.$

$f(21962)>0$, which says $f(x)\geq0$ for any $0<x<21962$ and it's enough to prove our inequality for $x>21962.$

Can you end it now?