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, $m+n=19$
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