# Expectation of random matrix

Assume $$Q$$ is a positive definite random matrix such that $$0 < \lambda_{\min}(Q)....\leq \lambda_{\max}(Q) \leq 1$$ holds. I want to show that \begin{align} E\left[\frac{\lambda_{\min}(Q)}{\lambda_{\min}(Q)+\lambda_{\max}(Q)}\right] \geq \frac{\lambda_{\min}(E[Q])}{\lambda_{\min}(E[Q])+\lambda_{\max}(E[Q])} \end{align}

Or I want to show the following Kantorovich type inequality for expected values: \begin{align} E\left[\frac{(y^Ty)^2}{(y^TQy)(y^TQ^{-1}y)}\right] \geq \frac{4\lambda_{\min}(E[Q])\lambda_{\max}(E[Q])}{\left[\lambda_{\min}(E[Q])+\lambda_{\max}(E[Q])\right]^2} \end{align}

$$\newcommand\lmax{\lambda_{\max}(P)}\newcommand\lmin{\lambda_{\min}(P)}$$Your first displayed inequality for all positive definite random matrices $$Q$$ means exactly that the function $$P\mapsto\frac\lmin{\lmin+\lmax}$$ on the set of all positive definite matrices is convex. Looking at just the diagonal positive definite matrices, we see that this convexity implies the convexity of $$\frac x{x+y}$$ in $$x,y>0$$ such that $$x\le y$$. However, $$\frac x{x+y}$$ is not convex in $$x\in(0,y]$$ for any $$y>0$$.
So, your first displayed inequality does not hold for some positive definite random matrices $$Q$$.
The second inequality is also false in general. E.g., let $$P(Q=D(1,a))=1/2=P(Q=D(a,1))$$, where $$0 and $$D(s,t)$$ stands for the diagonal $$2\times2$$ matrix with the diagonal entries $$s,t$$. Then $$EQ=D(\frac{1+a}2,\frac{1+a}2)$$ and hence the right-hand side of your second inequality is $$1$$. On the other hand, for any $$2\times1$$ matrix $$y$$ with both entries nonzero, the left-hand side of your second inequality goes to $$0$$ as $$a\downarrow0$$. So, your second inequality does not hold.