If $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $, then either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ For $a, b = 1, 2, 3, \cdots$, let $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $. Then prove that either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ holds.

Either $\frac{1}{2a}\leq \frac{1}{2b}$, in which case
$$\frac{1}{b}=\frac{1}{2b}+\frac{1}{2b}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2a}+\frac{1}{2a}=\frac{1}{a},$$
or $\frac{1}{2b}\leq \frac{1}{2a}$, in which case
$$\frac{1}{a}=\frac{1}{2a}+\frac{1}{2a}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2b}+\frac{1}{2b}=\frac{1}{b}.$$
 Now take reciprocals.
