# Prove that the variance of the Generalized Least squares estimator is less than the variance of the OLS estimator

Suppose, we consider the following regression model, $$Y = X\beta + \varepsilon$$ where $$\varepsilon$$ ~ $$N(0, \sigma^2V)$$ and V is a known $$n\times n$$ non-singular, positive definite square matrix.

Now OLS estimator of $$\beta$$ is $$\tilde \beta = (X'X)^{-1}X'Y$$ whereas, the GLS estimator of $$\beta$$ is $$\hat \beta = (X'V^{-1}X)^{-1}X'V^{-1}Y$$.

Then I need to show that the variance of GLS estimator $$V(\hat \beta)$$ is smaller than the variance of OLS estimator $$V(\tilde \beta)$$. I was able to find that $$V(\hat \beta) = \sigma^2(X'V^{-1}X)^{-1}$$ and $$V(\tilde \beta) = \sigma^2(X'X)^{-1}X'VX(X'X)^{-1}$$

So, I am trying to show that, \begin{align} V(\hat \beta) - V(\tilde \beta) & = \sigma^2(X'X)^{-1}X'VX(X'X)^{-1} - \sigma^2(X'V^{-1}X)^{-1} \end{align} this difference is a positive definite matrix. That's where I am stuck. I have no idea how to do that. Can anybody help??

please note that I know that in this case, the GLS estimator is BLUE (Best Linear Unbiased Estimator) according to Gauss-Markov Theorem. So It has the minimum variance. But I need to prove this particular case directly.