Trace inequality $tr(|XY|) \leq \|X\| tr(|Y|) $

Question

Why does one have $tr(|XY|) \leq \|X\| tr(|Y|) $ for any complex matrices? I do know that Cauchy Schwarz establishes $|tr(X^*Y)|\leq \|X\| \|Y\|$.

Ok so far I have $\langle u,X^*Xu \rangle=\langle Xu,Xu \rangle \leq \|X^2\| \langle u,u \rangle $ hence $X^*X\leq \|X^2\|$. This implies $Y^*X^*XY \leq Y^*\|X^2\|Y$. Then we have $tr(Y^*X^*XY)\leq tr(Y^*\|X^2\|Y)$ by linearity we are then done.