Showing that $||X||_1 = \max_{−I \leq H \leq I} \text{tr}[XH]$ where $||.||_1$ is the Schatten 1-norm

Question

Suppose $X$ is a Hermitia and define the schatten 1-norm as the sum of the absolute values of the eigenvalues of $X$. We can then decompose $X$ as $$X = X_+ − X_− $$ where $X_+$ is the projection of $X$ onto the eigenspace corresponding to all positive eigenvalues and $X_-$ is the projection of $X$ onto the eigenspace corresponding to all negative eigenvalues. Then we have that $X_+, X_−$ are positive semi-definite and $X_+X_− = X_−X_+ = 0$. Next, denote $P_−$ to be the projection to the negative eigenspace of $X$, and by $P_+ = I − P_−$ the projection to the positive eigenspace of $X$. Then $$|X| = H_+ + H_−,$$ so the Schatten 1-norm of $X$ can be written as $$||X||_1 = \text{tr}[X_+] + \text{tr}[X_−] = \text{tr}[XP_+] − \text{tr}[XP_−].$$ I am then trying to show that $$||X||_1 = \max_{−I \leq H \leq I} \text{tr}[XH]$$ where the maximum is taken over all Hermitian matrices H with eigenvalues between −1 and 1. I'm not quite sure where to go from here to show that $$||X||_1 \geq \text{tr}[XH]$$ for any $−I \leq H \leq I$.