$AD$ has exactly one negative eigenvalue if $v^T A v > 0$ and $D = \mbox{diag}(-1,1,1)$

Consider size $n \times n$ case with $D=\mbox{diag}(-1,1,\dots,1)$. As @user1551 write in his answer, $AD$ has at least one negative eigenvalue.

Suppose $\lambda \neq \eta$ is two negative eigenvalues of $AD$ with eigenvectors $v, w$, respectively; i.e. $ADv=\lambda v, ADw=\eta w$. Since $v$ and $w$ are linearly independent, so are $Dv$ and $Dw$. For every $s,t \in \mathbb R$ $sDv+tDw$ is nonzero unless $s^2+t^2=0$. It follows that $(sDv+tDw)^TA(sDv+tDw)>0$. Expand this yields $$ s^2(\lambda v^T Dv) + st(\lambda+\eta)v^T D w+ t^2 (\eta w^T D w) >0$$

Deduce that $v^TDv<0$ and $(w^TDw)(v^TDv)>(v^TDw)^2$. Define a symmetric matrix $B$ by $$B=Dvv^TD-(v^TDv)D$$

Then $Bv=0$. In other words, $v$ is an eigenvector of $B$ with eigenvalue $0$. Consider the subspace $U$ of $\mathbb R^n$, given by the intersection of the orthogonal complements of subspaces generated by $v$ and $e_1=(1,0,\dots,0)$; i.e. $U=\langle v \rangle^\perp \cap \langle e_1 \rangle ^\perp$. Check that $\dim U \geq n-2$. For all $u \in U$ we have $Bu = -(v^T D v)u $, because $v^Tu=0$ and $Du=u$. Finally, observe that $$\mbox{tr}(B)=v^Tv+(n-2)(-v^T Dv) $$

This shows that $B$ is positive semi-definite. Thus $$(v^TDw)^2-(w^TDw)(v^TDv)=(w^TDv)^2-(w^TDw)(v^TDv) = w^T B w \geq 0$$

contradiction.