(a)
Since, $H = \frac{2P\mathcal{Q}}{P + \mathcal{Q}}$
$\Rightarrow \frac{H}{P} = \frac{2\mathcal{Q}}{P + \mathcal{Q}}$
and $\frac{H}{\mathcal{Q}} = \frac{2P}{P + \mathcal{Q}}$
$\therefore\ \frac{H}{P} + \frac{H}{\mathcal{Q}} = \frac{2\mathcal{Q}}{P + \mathcal{Q}} + \frac{2P}{P + \mathcal{Q}} = 2$