IRL_ABooklet

To prove $(ABC)=\frac{a^2}{2(\cot B+\cot C)}$, we recall that $\sin (B+C)=\sin (A)$ because $\angle A+\angle B+\angle C=180^{\circ }$ and obtain $$\begin{array}{rl}\cot B+\cot C& =\frac{\cos B\sin C+\cos C\sin B}{\sin B\sin C}=\frac{\sin (B+C)}{\sin B\sin C}\\ & =\frac{\sin A}{\sin B\sin C}=\frac{a}{b\sin C}\text{(Sine Rule)}\\ & =\frac{a^2}{ab\sin C}=\frac{a^2}{2(ABC)},\end{array}$$ the required result.

Here is an alternative way to prove this formula. Let $D$ be the foot of the altitude from $A$, $x=|AD|$, and $y=|CD|$ where $x$ is taken negative if $\angle B$ is obtuse and $y$ is taken negative if $\angle C$ is obtuse. We then have $a=x+y$, $\cot B=x/h$ and $\cot C=y/h$, hence $\cot B+\cot C=a/h$. Using $2(ABC)=ah$ this turns into the desired formula.

To show that $\cos A\cos B\cos C\le \frac{1}{8}$ for all acute-angled triangles $ABC$, we use the area formula shown above. Since $B,C$ are acute angles, $\cot B$ and $\cot C$ are positive and we can use the AM-GM inequality to obtain $$\sqrt{\cot B\cot C}\le \frac{\cot B+\cot C}{2}=\frac{a^2}{4(ABC)},$$ with equality iff $\angle B=\angle C$. Whence $$\begin{array}{rl}(ABC)& \le \frac{a^2}{4}\sqrt{\tan B\tan C},\\ & =\frac{a}{2}\frac{\sqrt{\left(\frac{1}{2}ab\sin C\right)\left(\frac{1}{2}ac\sin B\right)}}{\sqrt{bc\cos B\cos C}}=\frac{a}{2}\frac{(ABC)}{\sqrt{bc\cos B\cos C}},\end{array}$$ and so $$\cos B\cos C\le \frac{a^2}{4bc},(8)$$ with equality iff $\angle B=\angle C$. Similarly, $$\cos C\cos A\le \frac{b^2}{4ca},\text{with equality iff }\angle C=\angle A\text{ and}$$ $$\cos A\cos B\le \frac{c^2}{4ab},\text{with equality iff }\angle A=\angle B.$$ Hence $$(\cos A\cos B\cos C{)}^2\le \frac{1}{64},$$ with equality iff $\angle A=\angle B=\angle C$, whence the desired inequality follows.

An alternative proof of inequality (8), not using the area formula we have shown in the first part, may use the Cosine Rule as follows: $$\begin{array}{rl}\cos B\cos C& =\frac{(a^2-b^2+c^2)}{2ac}\cdot \frac{(a^2+b^2-c^2)}{2ab}\\ & =\frac{a^4-(b^2-c^2{)}^2}{4a^{2}bc}\le \frac{a^4}{4a^{2}bc}=\frac{a^2}{4bc},\end{array}$$