A special sum involving the von Mangoldt function

Question

Assuming the Prime Number Theorem (if you need), show that $$ \sum_{a\leq X} \left|\sum_{b\leq X/a} (\Lambda(b) - 1) \right| = o(X\log X) $$ where $\Lambda$ is the von Mangoldt function.

The big issue in my tries is that $\sum_{a\leq X} \sum_{b\leq X/a} 1 = O(X\log X)$ and I want little $o$ instead. This $(-1)$ somehow has to play a big role in neutralizing enough of the contribution of $\Lambda$, but I cannot see how.

Any help appreciated!

Answer 1

$$\sum_{a \le X}o(X/a) = \sum_{a \le X/\log X} o(X/a) +O(\sum_{X/\log X<a \le X} X/a)$$ $$= o( X \log X) +O(X (\log X-\log (X/\log X)))= o(X\log X)$$