For IID $x$ drawn from $\mathcal{N}(0,\Sigma)$, is the following true?
$$\mathbb E \left[ \left\|\frac 1b \sum_{i=1}^b x_ix_i^T\right\|_2\right ]\le \frac{b+1}{b}\|\Sigma\|+\frac{1}{b}\operatorname{Tr}\Sigma$$
Simulations suggest it's true.
Formula comes by extrapolating from analogous result for expected Frobenius norm $$\mathbb E \left[ \left\|\frac 1b \sum_{i=1}^b x_ix_i^T\right\|_F^2\right ]=\frac{(b+1)}{b}\operatorname{Tr}\Sigma^2+\frac{1}{b} (\operatorname{Tr} \Sigma)^2$$