Aloha :)
Willkommen in der Mathelounge... \o/
Wir wissen, dass \(0<x\le y\) gilt. Daraus folgern wir:
$$x\le y\stackrel{(\cdot x)}{\implies}x^2\le xy\stackrel{(+xy)}{\implies}x^2+xy\le2xy\implies x(x+y)\le2xy$$$$\qquad\stackrel{:(x+y)}{\implies}x\le\frac{2xy}{x+y}\implies x^2\le\left(\frac{2xy}{x+y}\right)^2$$$$0\le(x-y)^2=x^2-2xy+y^2\stackrel{(+4xy)}{\implies}4xy\le x^2+2xy+y^2\implies 4xy\le(x+y)^2$$$$\qquad\stackrel{(:(x+y)^2)}{\implies}\frac{4xy}{(x+y)^2}\le1\stackrel{(\cdot xy)}{\implies}\frac{4x^2y^2}{(x+y)^2}\le xy\implies\left(\frac{2xy}{x+y}\right)^2\le xy$$