finding the shortest distance of a hermitian matrix to a set of hermitian matricies with specific eigenvalues 2-norm The title is more general, and all that I require is to show an inequality that I already have verified using random matrices in matlab.
Let $\lambda_1 \leq ... \leq \lambda$ and $\mu_1 \leq ... \leq \mu_n$ all in $\mathbb{R}$
let $X=(x_{ij}) \in \mathbb{C}^{n \times n}$ a unitary matrix ($\in U(n)$)
show:
$$\max_{|v|=1}|\langle ( diag(\lambda_1,...,\lambda_n) -X^*diag(\mu_1,...,\mu_n)X)v , v \rangle| \geq \max_{i=1,...,n}|\lambda_i - \mu_i| $$
or equivalently show:
$$ \text{absolute maximum eigenvalue of } \\
diag(\lambda_1,...,\lambda_n) -X^*diag(\mu_1,...,\mu_n)X \text{ is } \geq \max_{i=1,...,n}|\lambda_i - \mu_i|$$
Equality happens if for example $X=Id$. I would be happy enough to show this for $\mathbb{R}$ and orthogonal matrices $O(n)$ instead of $\mathbb{C}$ and $U(n)$. 
Background:
this is the final step for showing that if a hermitian matrix $M$ has eigenvalues $\mu_i$ to eigenvectors $w_i, \ W=(w_1,...,w_n)$, then the closest hermitian matrix to $M$ with given eigenvalues $\lambda_i$ is $W^*diag(\lambda_1,...,\lambda_n)W$.
For n = 3 to 6 I have generated random eigenvalues and then 100.000 random orthogonal matrices in matlab. The computations confirm this inequality. Please help proving it. I gladly cite the solution where I need it and share the link to the finished work here.
thank you!

Here is the answer:
https://arxiv.org/abs/1703.00829
Check the eigenvalue inequality with it's proof. It is based on Weyl's inequality.
