Here $x_1>x_2>0 \Rightarrow x_1 \ne x_2 \Rightarrow (x_1-x_2)^2 \ne 0$.
and we know from simple algebra that
$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2\tag{i}$
Now $\forall~ x_1, x_2 \in \mathbb{R}$ and $x_1 \ne x_2$, we obtain
$$\begin{align} (x_1-x_2)^2 &>0 \\ \Rightarrow (x_1+x_2)^2-4x_1x_2 &>0 ~;~[\mathrm{From~no(i)}] \\ \Rightarrow (x_1+x_2)^2 &> 4x_1x_2 \\ \Rightarrow x_1+x_2 &>2\cdot\sqrt{x_1x_2}\\ \Rightarrow \frac{x_1+x_2}{2}&> (x_1x_2)^{\frac{1}{2}} \\ \Rightarrow \log\left( \frac{x_1+x_2}{2} \right) &> \log(x_1x_2)^{\frac{1}{2}} \\ \Rightarrow \log\left( \frac{x_1+x_2}{2} \right) &> \frac{1}{2}\log(x_1x_2)~;~[\because \log a^r=r\log a] \\ \Rightarrow \log\left( \frac{x_1+x_2}{2} \right) &> \frac{\log(x_1)+\log(x_2)}{2} ~;~[\because \log (xy)=\log x + \log y] \end{align} $$
So the correct answer is A.