$A\ge t+\frac{4}{t} =f(t) (t=x+y \le \frac{4}{3})$$f'(t)=1-\frac{4}{t^2}=0\Leftrightarrow x=\pm2$$\forall0\le t\le \frac{4}{3}\Rightarrow f(t)\ge f(\frac{4}{3})=\frac{13}{3}$
$x+y\ge 2\sqrt{xy};\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}$$\Rightarrow (x+y)(\frac{1}{x}+\frac{1}{y})\ge2.2=4\Rightarrow \frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}$$A\ge t+\frac{4}{t} =f(t) (t=x+y \le \frac{4}{3})$$f'(t)=1-\frac{4}{t^2}=0\Leftrightarrow x=\pm2$$\forall0\le t\le \frac{4}{3}\Rightarrow f(t)\ge f(\frac{4}{3})=\frac{13}{3}$