Using A.M. $\geq$ H.M.

$\Rightarrow \frac{a+b+c}{3} \geq \frac{3}{\frac 1a +\frac 1b+\frac 1c}$

$\Rightarrow (a+b+c)(\frac 1a +\frac 1b+\frac 1c) \geq 9$

$\Rightarrow (a+b+c)^2 \geq 9$

Let $A=a+b+c$

then $A^2 \geq 9 \Rightarrow \frac{1}{A^2} \leq 9$

let, $f(x)=\frac{1}{(A+x)^2}$

$\lambda_i = \frac 13$

$x_1=a,x_2=b,x_3=c$

Function is concave.

using JENSEN's inequality :

$\displaystyle\sum_{i=1}^n \lambda_i f(x_i) \leq f\bigg(\lambda_i x_1\bigg)$

$\Rightarrow   \frac{\text{Required Expression}}{3}  \leq f\bigg(\frac A3\bigg)$

$\Rightarrow   \frac{\text{Required Expression}}{3}  \leq \frac{1}{(A+\frac A3)^2}$

$\Rightarrow  \frac{\text{Required Expression}}{3}  \leq \frac{9}{16A^2} \leq \frac {1}{16} $

$\Rightarrow \text{ Required Expression } \leq \frac {3}{16} =\frac mn$

Hence, $m+n=19$