Consider the simple case of inference for the mean (in other words, I have not figured out the general case yet :-)), aka regression on a constant. Take $\sigma^2=1$ for simplicity.

Then, $V(\hat\beta_A)=1/n$, $V(\hat\beta_B)=1/m$, $V(\hat\beta)=1/(n+m)$. Write $T=n+m$ and $n=cT$, $m=(1-c)T$, $c\in(0,1)$. Then, and there is a little mistake in what you write, as constants leave variances in squares, $$ V(\tilde\beta)=\frac{1}{4}\left(\frac{1}{cT}+\frac{1}{(1-c)T}\right)=\frac{1}{4}\left(\frac{T}{cT(1-c)T}\right)=\frac{1}{4}\left(\frac{1}{c(1-c)T}\right) $$ We have that $c(1-c)$ is largest at $c=1/2$, so that $$ V(\tilde\beta)\geq\frac{1}{4}\left(\frac{1}{0.5(1-0.5)T}\right)=1/T=Var(\hat\beta) $$

Consider the simple case of inference for the mean (see whuber's answer or the ideas from the comments for the general case), aka regression on a constant. Take $\sigma^2=1$ for simplicity.

Then, $V(\hat\beta_A)=1/n$, $V(\hat\beta_B)=1/m$, $V(\hat\beta)=1/(n+m)$. Write $T=n+m$ and $n=cT$, $m=(1-c)T$, $c\in(0,1)$. Then, and there is a little mistake in what you write, as constants leave variances in squares, $$ V(\tilde\beta)=\frac{1}{4}\left(\frac{1}{cT}+\frac{1}{(1-c)T}\right)=\frac{1}{4}\left(\frac{T}{cT(1-c)T}\right)=\frac{1}{4}\left(\frac{1}{c(1-c)T}\right) $$ We have that $c(1-c)$ is largest at $c=1/2$, so that $$ V(\tilde\beta)\geq\frac{1}{4}\left(\frac{1}{0.5(1-0.5)T}\right)=1/T=Var(\hat\beta) $$