Inequality 1

Method 1:

By AM-GM, $\frac{u+v}{2} + \frac{v+w}{2} + \frac{w+u}{2} \geq \frac{3}{2}\sqrt[3]{(u+v)(v+w)(w+u)}$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{3}{\sqrt[3]{xyz}}$ $\frac{x}{u+v}+\frac{y}{v+w}+\frac{z}{w+u} \geq 3\sqrt[3]\frac{xyz}{(u+v)(v+w)(w+u)}$

Multiply the three inequalities together to yield $(u+v+w)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{x}{u+v}+\frac{y}{v+w}+\frac{z}{w+u}\right)\geq \frac{27}{2}$

$m-n = 27-2 = 25$

Method 2:

By Holder's Inequality, $\frac12\sum_{cyc}(u+v) \cdot \sum_{cyc} \frac1x \cdot \sum_{cyc} \frac{x}{u+v} \ge \frac{3^3}2$

And the result follows.