I think it means that we need to prove that:
$$\sqrt{(x^xy^y)^{\frac{1}{x+y}}\sqrt{xy}}\leq \frac{x+y}{2},$$
otherwise, your inequality is wrong for $x=y=1$.
Let $y=kx$.
Since our inequality is symmetric, we can assume $k\geq1.$
Thus, we need to prove that
$$x^xy^y\leq \left(\frac{(x+y)^2}{4\sqrt{xy}}\right)^{x+y}$$ or
$$x^x(kx)^{kx}\leq\left(\frac{x(1+k)^2}{4\sqrt{k}}\right)^{x(1+k)}$$ or
$$x(kx)^k\leq\left(\frac{x(1+k)^2}{4\sqrt{k}}\right)^{1+k}$$ or
$$k^k\leq \left(\frac{(1+k)^2}{4\sqrt{k}}\right)^{1+k}$$ or $f(k)\geq0,$ where
$$f(k)=2(1+k)\ln(1+k)-k\ln{k}-\frac{1+k}{2}\ln{k}-(k+1)\ln4.$$
Now, we obtain:
$$f'(k)=2\ln(1+k)+2-\ln{k}-1-\frac{1}{2}\ln{k}-\frac{1+k}{2k}-\ln4,$$
$$f''(k)=\frac{2}{1+k}-\frac{3}{2k}+\frac{1}{2k^2}=\frac{(k-1)^2}{2k^2(1+k)}\geq0,$$ which gives
$$f'(k)\geq f(1)=0$$ and from here
$$f(k)\geq f(1)=0$$ and we are done!