Prove simple upper and lower bounds on logarithm Prove or disprove that, for $x,y>1$:
$$
\frac{1}{2} \log(x y) \le \log(x + y) \le \log(x y)
$$
At first, I thought the right inequality follows since $x+y < xy$ and $\log$ is monotonically increasing, but actually this doesn't hold for all $x,y>1$.

For the first part you have that
$$
\frac{1}{2}\log \left( {xy} \right) = \log \left( {\sqrt {xy} } \right) \le \log \left( {\frac{{x + y}}{2}} \right) \le \log \left( {x + y} \right)
$$
The inequality $$
\log \left( {\sqrt {xy} } \right) \le \log \left( {\frac{{x + y}}{2}} \right)
$$
follows from the fact that
$$
\sqrt {xy}  \le \frac{{x + y}}{2}
$$
for each $x>0, y>0$ and from the fact that the function t\to \log t$ is monotonically increasing. The second part of your statement is trivially false as already pointed out.
