Is either of these inequalities true?
$$\lambda(tA + (1-t)B)\geq t\lambda(A) + (1-t)\lambda(B)$$
or
$$\lambda(tA + (1-t)B)\leq t\lambda(A) + (1-t)\lambda(B),$$
where $0\leq t \leq 1$, $A,B$ are bounded domains in $\mathbb{R}^n$ and $\lambda(\Omega)$ is an eigenvalue of the problem
$$-\Delta\,u=\lambda\,u\,\,\mbox{in}\,\,\, \Omega, \, u=0\,\,\,\mbox{on}\,\,\, \partial\Omega.$$



An inequality for eigenvalues of the Dirichlet problem
**Update:** I intended this to be a complete answer originally, but my "counterexample" was based on a miscalculation. So this is now more a collection of remarks on what I think the question is about.


We need to be more specific about which eigenvalue we want to take. I will discuss the ground state energy, which seems natural.


The first inequality is clearly false because $\lambda(\Omega)$ becomes large when we make the region small. As for the second, I want to rename $A'=tA$, $B'=(1-t)B$ (and then drop the primes again). Then the claim becomes
$$
\lambda(A+B) \le t^3\lambda (A) + (1-t)^3 \lambda(B) ,
$$
for all $0\le t\le 1$.
Minimize over $t$ to rewrite this as
$$
\lambda(A+B) \le \frac{\lambda(A)\lambda(B)}{(\sqrt{\lambda (A)} + \sqrt{\lambda (B)})^2} . \quad\quad\quad\quad (1)
$$
This is true in one dimension, with equality; recall that $\lambda([0,L])=\pi^2/L^2$ to see this. I don't know what happens in higher dimensions; preliminary attempts at easy counterexamples (rectangles ...) were unsuccessful, and if I had to guess, I would now say that (1) is probably true. The inequality does hold in the special case $A=B$ (because then $A+A\supseteq 2A$).


