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}



Expectation of random matrix
$\newcommand\lmax{\lambda\_{\max}(P)}\newcommand\lmin{\lambda\_{\min}(P)}$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 $0<a<1$ 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.


