Aloha :)
Für \(x,y>0\) gilt:$$0\le(\sqrt x-\sqrt y\,)^2=(\,\sqrt x\,)^2-2\sqrt{xy}+(\,\sqrt y\,)^2=x-2\sqrt{xy}+y$$$$\implies\;2\sqrt{xy}\le x+y\;\implies\;\sqrt{xy}\le\frac{x+y}{2}$$Damit sind wir fertig, denn:
$$\ln\left(\frac{x+y}{2}\right)\ge\ln\left(\sqrt{xy}\right)=\ln\left((xy)^{\frac{1}{2}}\right)=\frac{1}{2}\ln(xy)=\frac{\ln(xy)}{2}=\frac{\ln(x)+\ln(y)}{2}$$