Let $a, c > 0$ and $b \in \mathbb{R}$. Show that for all $(x,y) \in \mathbb{R}^2$,
\begin{equation*}a x^2 - 2bxy + cy^2 \ge \frac{ac - b^2}{2}\min(a^{-1}, c^{-1}) (x^2 + y^2).\end{equation*}This lower bound is asserted on page 1842 of the article Banks-Kojima 2002 (appearing in the journal Inverse Problems). It is stated that the proof of the inequality is elementary but no proof is given.
For starters let us consider the case when $a = c$. Then we need to show
\begin{equation*}\frac{a^2}{2} x^2 - 2abxy + \frac{a^2}{2}y^2 \ge -\frac{b^2}{2} (x^2 + y^2),\end{equation*}which follows immediately from applying Young's inequality a few times,\begin{equation*}|2abxy| \le (\frac{a^2}{2} + \frac{b^2}{2}) 2|xy| \le (\frac{a^2}{2} + \frac{b^2}{2})(x^2 + y^2).\end{equation*}The case $a \neq c$ seems more difficult. For instance, if $a \ge c$, we need to see that\begin{equation*}a^2 x^2 - 2abxy + cay^2 + \frac{b^2}{2}(x^2 + y^2) \ge \frac{ac}{2} (x^2 + y^2).\end{equation*}It is no longer clear to me how to go about bounding $|2abxy|$.
Edit: From an answer below and its comments, it is clear that the estimate holds if $ac - b^2 \ge 0$, but can fail in certain situations where $ac - b^2 < 0$.