Prove $ac\cos B+ab\cos C-bc\cos A-a^2 \le \frac{c^2}{8\cos^2(90^\circ-C)}$ for $\triangle ABC$

Question

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 1

Id est, by the law of cosines we need to prove that: $$\frac{a^2+c^2-b^2}{2}+\frac{a^2+b^2-c^2}{2}-\frac{b^2+c^2-a^2}{2}-a^2\leq\frac{c^2}{8\left(\frac{2S}{ab}\right)^2},$$ where $S=\frac{1}{4}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}$.

Thus, we need to prove that $$b^2+c^2-a^2+\frac{a^2b^2c^2}{\sum\limits_{cyc}(2a^2b^2-a^4)}\geq0.$$ Now, let $a=y+z$, $b=x+z$ and $c=x+y$.

Thus, $x$, $y$ and $z$ are positives and we need to prove that: $$2(x^2+xy+xz-yz)+\frac{\prod\limits_{cyc}(x+y)^2}{16xyz(x+y+z)}$$ or $$(y^2+34yz+z^2)x^4+2(y^3+35yz+35y^2z^2+z^4)x^3+$$ $$+(y^4+38y^3z+42y^2z^2+38yz^3+z^4)x^2+$$ $$+2yz(y^3-13y^2z-13yz^2+z^3)x+y^2z^2(y+z)^2\geq0.$$ Now, let $x^4=t\cdot\frac{y^2z^2(y^2+10yz+z^2)}{12}.$

Thus, since $$y^2+34yz+z^2\geq36\sqrt[3]{\frac{y^2z^2(y^2+10yz+z^2)}{12}},$$ $$2(y^3+35y^2z+35yz^2+z^3)\geq144\sqrt{\frac{y^2z^2(y^2+10yz+z^2)}{12}},$$ $$y^4+38y^3z+42y^2z^2+38yz^3+z^4\geq120\sqrt[3]{\left(\frac{y^2z^2(y^2+10yz+z^2)}{12}\right)^2},$$ $$2yz(y^3-13y^2z-13yz^2+z^3)\geq-48\sqrt[6]{\left(\frac{y^2z^2(y^2+10yz+z^2)}{12}\right)^5}$$ and $$y^2z^2(y+z)^2\geq4\cdot\frac{y^2z^2(y^2+10yz+z^2)}{12},$$ it's enough to prove that: $$36t^4+144t^3+120t^2-48t+4\geq0$$ or $$(3t^2+6t-1)^2\geq0$$ and we are done!

The equality occurs for $t=\frac{2}{\sqrt3}-1$ and, for example, for $y=z=1$, which gives $x=\frac{2}{\sqrt3}-1$ and we got a triangle with measured angles $30^{\circ}$, $30^{\circ}$ and $120^{\circ}.$

Answer 2

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 $\alpha=120^\circ$.