Bounds on expectation of Gaussian random vectors

Very nice questions, can you share where you got them from?

For the second one:

We can show this claim by showing that the right side is maximized for $D = I$.

$$\frac{E ||\sqrt D Y||_2}{\sqrt{tr(D)}}=E||\sqrt{\frac{D}{tr(D)}}Y||_2$$

The expectation of the norm and the expectation of the norm squared are both maximized by the same $D$ since squaring is monotonic for positive values. So we can instead maximize:

$$E||\sqrt{\frac{D}{tr(D)}}Y||_2^2$$

It's not too hard to show that this expression is convex is $D$ by showing the hessian is positive semidefinite. We can then show that the maximum is attained for $D = I$ by showing that the gradient is $0$ at that point:

$$\nabla_D E||\sqrt{\frac{D}{tr(D)}}Y||_2^2=E[\frac{YY^T}{tr(D)} - \frac{Y^TDY}{tr(D)^2}I]$$

Applying $E[YY^T] = I$ and $E[Y^TDY] = tr(D)$ shows that this is $0$ for $D = I$, and thus the claim holds.

I also tried another way by expressing the ratio in terms of the SVD of $D$. Then you end up with something like this for the ratio on the left:

$$E ||\sum_i \alpha_i \sqrt{\frac{\lambda_i}{\sum_j \lambda_j}}v_i||$$

where $\alpha_i$ is the length of the projection of $Y$ onto the eigenvector $v_i$ of $D$. And on the right you get:

$$E ||\sum_i \alpha_i \frac{1}{\sqrt m}v_i||$$

You can then show that the first expression is maximized when all $\lambda_i$ are equal. wlog, since $V$ (the matrix of eigenvectors) is orthonormal, this shows that this expression is maximized for $D=I$.