Question

For $x,y$ such that $x>y>0,xy=4$, find the minimum value of the expression $E=\frac{(x+y{)}^2}{x-y}$.

Answer 1

Answer: 0008

$E=\frac{(x-y{)}^2+4xy}{x-y}=\frac{(x-y{)}^2+16}{x-y}=(x-y)+\frac{16}{x-y}$
$={\left(\sqrt{x-y}-\frac{4}{\sqrt{x-y}}\right)}^2+8$
$\therefore E\ge 8$
$\therefore E_{\min }=8$.