Suggestion: We have the series expansion
$$
\arctan(\pi/z)/\pi=\sum_{k=0}^\infty (-1)^k \frac{\pi^{2k}}{(2k+1)z^{2k+1}}
$$
with $z$ respectively $\log(x)$, $\log(y)$, and $\log(x+y)$.  This is an alternating series with terms tending to $0$.  Since the partial sums alternately overestimates and underestimates the sum, it would suffice to prove the simpler inequality
$$
\frac{\pi^2}{3\log(x+y)^3}\le\frac{\pi^2}{3\log(x)^3}+\frac{\pi^2}{3\log(y)^3}-\frac{\pi^4}{5\log(x)^5}-\frac{\pi^4}{5\log(y)^5}.
$$


----------

Update:  For $y\ge 12$,
$$
\frac{\pi^2}{3\log(y)^3}-\frac{\pi^4}{5\log(y)^5}>0,
$$
and for any fixed $x$, as $y\to\infty$
$$
\frac{\pi^2}{3\log(x+y)^3}\to 0,
$$
so certainly as $y\to\infty$
$$
\frac{\pi^2}{3\log(x+y)^3}\le\frac{\pi^2}{3\log(x)^3}-\frac{\pi^4}{5\log(x)^5}.
$$
Thus the desired inequality holds in the region  above some curve, $y\ge y(x)$ (and the same reversing the roles of $x$ and $y$.)