Help: Partitioned samples efficiency in OLS compared to one sample regression As usual, we can estimate by OLS the model (in matrix form) $Y=\alpha+\beta*X+u$ with a sample of $n+m$ observations. The OLS estimator is $\hat{\beta}=(X^{T}X)^{-1}X^{T}Y$. Now, if we partition our sample in two subsamples A and B of sizes n and m, respectively, we could get the OLS estimators for each sumbsample: $\hat{\beta_A}=(X_A^{T}X_A)^{-1}X_A^{T}Y_A$ and $\hat{\beta_B}=(X_B^{T}X_B)^{-1}X_B^{T}Y_B$. We know both of these estimators are unbiased, and their variance is (assuming homoscedasticity) equal to $\sigma^2(X_i^{T}X_i)^{-1}$. Now let's define $\tilde{\beta}=\frac{\hat{\beta_A}+\hat{\beta_B}}{2}$, which is also unbiased, and, assuming no serial correlation, its variance is $V(\tilde{\beta})=\frac{V(\hat{\beta_A})+V(\hat{\beta_B})}{2}$.
How could we prove this estimator $\tilde{\beta}$ is less efficient/worse than $\hat{\beta}$?

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
$$
V(\tilde\beta)\geq\frac{1}{4}\left(\frac{1}{0.5(1-0.5)T}\right)=1/T=Var(\hat\beta)
$$
