$\newcommand\lmax{\lambda_{\max}(P)}\newcommand\lmin{\lambda_{\min}(P)}$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 nonzero $2\times1$ matrix $y$, the left-hand side of your second inequality goes to $0$ as $a\downarrow0$. So, your second inequality does not hold.

$\newcommand\lmax{\lambda_{\max}(P)}\newcommand\lmin{\lambda_{\min}(P)}$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.

$\newcommand\lmax{\lambda_{\max}(P)}\newcommand\lmin{\lambda_{\min}(P)}$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 $Q$.

$\newcommand\lmax{\lambda_{\max}(P)}\newcommand\lmin{\lambda_{\min}(P)}$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 nonzero $2\times1$ matrix $y$, the left-hand side of your second inequality goes to $0$ as $a\downarrow0$. So, your second inequality does not hold.