For a convex, analytic function $f: \mathbb{R} \to \mathbb{R}$ and a real-valued, symmetric matrix $M \in \mathbb{R}^{n \times n}$ we have the bound
$$\operatorname{Tr} f(M) \geq \sum_{i = 1}^n f (M_{ii}).$$
Here $f(M)$ denotes the matrix function of $f$ applied to $M$, not the element-wise application of $f$ to $M$.
Can the trace of $f(M)$ be similarly upper bounded by the trace of the element-wise application of $f$ to $M$? In particular I am interested in a bound for $f(x) = x^{-1}$ applied to positive-definite matrices.
Here is a counterexample for $f(x) = x^{-1}$, which is mentioned as of particual interest, but it does not fit to the assumptions on $f$ in the question.
Define for $\epsilon>0$ $$ A = \pmatrix{ 1 & 1 \\ 1 & 1 + \epsilon}, $$ which is positive definite. Its inverse is $$ A^{-1} = \epsilon^{-1}\pmatrix{ 1+ \epsilon &- 1 \\ -1 & 1 }, $$ with trace $tr(A^{-1}) = \frac{2+\epsilon}\epsilon$, which explodes for $\epsilon\searrow0$. However, the sum of the inverses of the diagonal elements of $A$ is $1 + \frac1{1+\epsilon}$, which is bounded.
Here is a counterexample for $f(x) = \exp(x)$. Define $$ A = \pmatrix{ 2t & t \\ t & 2t}, $$ with $t>0$. It has eigenvalues $t$ and $3t$. Then $tr(f(A)) = \exp(t) + \exp(3t)$, while $\sum_i f(a_{ii}) = 2\exp(2t)$. And $$ \frac{ tr(f(A)) }{ \sum_i f(a_{ii}) } = \frac12 (\exp(-t) + \exp(+t)) $$ which is unbounded for $t\to +\infty$.