When $a$ and $b$ are positive numbers and $k\in\mathbb{N}$, one has the bound $$ (a+b)^{2k} \leq 2^{2k} (a^{2k} + b^{2k}), $$ which can be proven for instance by bounding $a+b \leq \max(a,b)$.
I am curious whether a similar bound holds for psd (symmetric and positive semi-definite) matrices $A$ and $B$. That is, does it hold that $$ (A+B)^{2k} \leq 2^{2k} (A^{2k} + B^{2k})? $$
In the case for example that $k=1$ I believe one can modify the proof of the Cauchy-Schwartz inequality: $$ 2(A^2 + B^2) - (A+B)^2 = A^2 - AB - BA + B^2 = (A-B)^2. $$ The matrix on the right is positive semidefinite because it is a square, and therefore $$ 0 \leq 2(A^2 + B^2) - (A+B)^2, $$ which can be rearranged to $$ (A+B)^2 \leq 2 (A^2 + B^2). $$
I believe an induction can be done in the case that $k$ is a power of $2$, but I am curious if this bound holds for arbitrary $k$, ideally avoiding induction (just because I am interested in the shortest possible proof, if it exists).