The last term in the equation can be written as
$$
(X\beta - X\hat{\beta})'H^{-1}(X\beta - X\hat{\beta}).
$$
In this form the equation is saying something interesting. Assuming $H$ is positive definite and symmetric, so is its inverse. Therefore,we can define an inner product $<x, y>_{H^{-1}} = x'H^{-1}y$, giving us geometry. Then the above equality is essentially saying that,
$$
(X\beta - X\hat{\beta}) \perp (y - X\hat{\beta}).
$$
I wanted to give you this bit of intuition since a commenter has already left a link to the derivation.
Edit: For Posterity
LHS:
\begin{eqnarray}
(y-X \beta)'H^{-1}(y-X \beta) &=& y'H^{-1}y &-& 2y'H^{-1}X \beta &+& \beta'X'H^{-1}X\beta \\
&=& (A) &-& (B) &+& (C)
\end{eqnarray}
RHS:
$$
(y-X\hat\beta)'H^{-1}(y-X\hat\beta)+(\beta-\hat\beta)'(X'H^{-1}X)(\beta-\hat\beta)
$$
\begin{eqnarray}
&=& y'H^{-1}y &- 2y'H^{-1}X\hat{\beta} &+ \hat{\beta}'X'H^{-1}X\hat{\beta} &+ \beta X'H^{-1}X\beta &- 2\hat{\beta}X'H^{-1}X\beta &+ \hat{\beta}'X'H^{-1}X\hat{\beta} \\
&=& (A) &- (D) &+ (E) &+ (C) &- (F) &+ (E)
\end{eqnarray}
Relation:
$$
\hat{\beta} = (X'H^{-1}X)^{-1}X'H^{-1}y
$$
By plugging in the relation you can show that (B) = (F), and that 2(E) = (D). All done.