(i) Since $x$, $y$ are positive real numbers, we have:
$$\frac{4}{x+y} \le \frac{1}{x} + \frac{1}{y} \Leftrightarrow \frac{4}{x+y} \le \frac{x+y}{xy} \Leftrightarrow (x+y)^2 \ge 4xy \Leftrightarrow (x-y)^2 \ge 0.$$

(ii)
Let
$$x = (\alpha + \gamma)(\beta + \delta) + 4\alpha\gamma, \quad y = (\alpha + \gamma)(\beta + \delta) + 4\beta\delta,$$
and thus:
$$\begin{aligned} x + y &= 2(\alpha + \gamma)(\beta + \delta) + 4(\alpha\gamma + \beta\delta) \\ (\alpha + \beta)(\gamma + \delta) + (\beta + \gamma)(\alpha + \delta) &= \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \alpha\beta + \beta\delta + \alpha\gamma + \gamma\delta \\ &= (\alpha\beta + \alpha\delta + \beta\gamma + \gamma\delta) + 2(\alpha\gamma + \beta\delta) \\ &= (\alpha + \gamma)(\beta + \delta) + 2(\alpha\gamma + \beta\delta) = \frac{x+y}{2} \end{aligned}$$
Therefore the inequality becomes
$$\frac{4}{x+y} \le \frac{1}{x} + \frac{1}{y},$$
which is valid because of (i).

The equality holds if and only if $x = y \Leftrightarrow x - y = 0 \Leftrightarrow \alpha\gamma = \beta\delta$.