proof of Schur’s inequality

By Schur’s theorem, a unitary matrix $U$ and an upper triangular matrix $T$ exist such that $A=UT{U}^H$, $T$ being diagonal if and only if $A$ is normal.Then $A^{H}A=U{T}^H{U}^{H}UT{U}^H=U{T}^{H}T{U}^H$, which means $A^{H}A$ and $T^{H}T$ are similar; so they have the same trace. We have:

If and only if $A$ is normal, $T=D$ and therefore equality holds.$\mathrm{\square }$