For $\triangle ABC$, show that $ac\cos B+ab\cos C-bc\cos A-a^2 \le \frac{c^2}{8\cos^2(90^\circ-C)}$ 
Triangle $\triangle ABC$ has sides $a$, $b$, and $c$, and circumradius $R$. Prove that
$$ac \cos B + ab \cos C - bc \cos A - a^2 \le \frac{c^2}{8\cos^2(90^\circ - C)}$$
When does equality occur?

I came across this question in a different forum and I thought it was interesting. I made a bit of progress but not much: I changed $R^2$ to the fraction in the inequality. I think that there is probably another use of Law of Sines or Law of Cosines but I can't find one.
Edit: A lot of people have questions about if the problem is right; here is the original problem:
Triangle $\triangle ABC$ has sides $a$, $b$, and $c$, and circumradius $R$. Prove that
$b^2 + c^2 - a^2 \ge -R^2$
When does equality occur?

Answer to the second question (equality).

Triangle $ABC$ has sides $a$, $b$, and $c$, corresponding angles
$\alpha,\beta,\gamma$,  semiperimeter $\rho$, inradius $r$ and
circumradius $R$.  Prove that
\begin{align} R^2-a^2+b^2+c^2\ge0\tag{1}\label{1}. \end{align}
When does equality occur?

By dividing \eqref{1} by $R^2$, we have
\begin{align}
1-4\sin^2\alpha+4\sin^2\beta+4\sin^2\gamma&\ge0
\tag{2}\label{2}
.
\end{align}
It's easy to verify that \eqref{2} becomes an equality
for $\alpha=120^\circ,\beta=\gamma=30^\circ$.
In other words,
\eqref{1} becomes an equality
for an isosceles triangle with $\alpha=120^\circ$.
