Selection Examination

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

(ii) Let $$x=(\alpha +\gamma )(\beta +\delta )+4\alpha \gamma ,y=(\alpha +\gamma )(\beta +\delta )+4\beta \delta ,$$ and thus: $$\begin{array}{rl}x+y& =2(\alpha +\gamma )(\beta +\delta )+4(\alpha \gamma +\beta \delta )\\ (\alpha +\beta )(\gamma +\delta )+(\beta +\gamma )(\alpha +\delta )& =\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\alpha \beta +\beta \delta +\alpha \gamma +\gamma \delta \\ & =(\alpha \beta +\alpha \delta +\beta \gamma +\gamma \delta )+2(\alpha \gamma +\beta \delta )\\ & =(\alpha +\gamma )(\beta +\delta )+2(\alpha \gamma +\beta \delta )=\frac{x+y}{2}\end{array}$$ 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 $x=y\iff x-y=0\iff \alpha \gamma =\beta \delta$.