An algebra problem by Victor Loh

By Cauchy-Schwarz, we have

$\left[(b+c)+(a+c)+(a+b)\right]\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq 9,$

or

$2\left(\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\right)\geq 9,$

which yields the desired result

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}$

so $m+n=\boxed{5}$ and we are done.

using AM - HM inequality

By A.M.-G.M. inequality

$\frac {a+b}{2}$ $\geq$ $\sqrt{ab}$ $\Rightarrow$ a+b $\geq$ 2 $\sqrt{ab}$ ----->(1)

$\frac {b+c}{2}$ $\geq$ $\sqrt{bc}$ $\Rightarrow$ b+c $\geq$ 2 $\sqrt{bc}$ ----->(2)

$\frac {c+a}{2}$ $\geq$ $\sqrt{ca}$ $\Rightarrow$ c+a $\geq$ 2 $\sqrt{ca}$ ----->(3)

$\frac {a}{b+c}$ + $\frac {b}{c+a}$ + $\frac {c}{a+b}$ $\geq$ N

$\Rightarrow$ $\frac {a}{2\sqrt{bc}}$ + $\frac {b}{2\sqrt{ca}}$ + $\frac {c}{2\sqrt{ab}}$ $\geq$ N

$\Rightarrow$ $\frac {2a^{2}\sqrt{bc}+2b^{2}\sqrt{ca}+2c^{2}\sqrt{ab}}{8abc}$ $\geq$ N

$\Rightarrow$ $\frac {2abc(\frac {a}{\sqrt{bc}}+\frac {b}{\sqrt{ca}}+\frac {c}{\sqrt{ab}})}{8abc}$ $\geq$ N

Now

$\frac {a}{\sqrt{bc}}$ + $\frac {b}{ca}$ + $\frac {c}{ab}$ $\geq$ N $\Rightarrow$ $\frac {a}{\sqrt{bc}}$ + $\frac {b}{ca}$ + $\frac {c}{ab}$ $\geq$ 4

$\Rightarrow$ N=4

We can express 4 in the form of $\frac {m}{n}$ as $\frac {4}{1}$ $\Rightarrow$ m+n=4+1= $\boxed{5}$