There are probably many ways of doing it. Here is one:
Note first that $A$ admits a square root, i.e. there is a symmetric positive definite matrix $A^{1/2}$ such that $(A^{1/2})^2=A$. This is done by writing $A$, via the Spectral Theorem, as
$$
A=\sum_{j=1}^n \lambda_j P_j,
$$
where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ counting multiplicities, and $P_1,\ldots,P_n$ are pairwise orthogonal projections of rank one (i.e. the projections onto the corresponding eigenspaces). Then define
$$
A^{1/2}=\sum_{j=1}^n \lambda_j^{1/2} P_j.
$$
Now, since $L$ is invertible, we can write
$$
D=MAM^T,
$$
where $M=L^{-1}$.
Considering the canonical basis $\{e_1,\ldots,e_n\}$, we have
$$
D_{kk}=\langle De_k,e_k\rangle = \langle MAM^Te_k,e_k\rangle
=\langle A^{1/2}M^Te_k, A^{1/2}M^Te_k\rangle\geq0.
$$
But $D$ is invertible, so $D_{kk}\ne0$, and so $D_{kk}>0$ for all $k=1.\ldots,n$.