# triangle inequality with $\cos$ and triangle angles

In any triangle $$ABC$$

If: $$\frac{1}{8}\geq \cos A\cdot \cos B\cdot \cos C > y$$, find the value of $$y$$.

## 1 Answer

Find the minimal value of \begin{align} t&=\cos A\cdot \cos B\cdot \cos C \tag{1}\label{1} , \end{align}

where $$A,B,C$$ are the angles of $$\triangle ABC$$.

For any $$\triangle ABC$$ with semiperimeter $$\rho$$ inradius $$r$$ and circumradius $$R$$, we can using a known expression for \eqref{1} in terms of two parameters $$v=r/R$$, $$v\in[0,\tfrac12]$$, and $$u=\rho/R=u(v)$$, for any valid value of $$v$$ $$u(v)\in[u_{\min}(v),\, u_{\max}(v)]$$:

\begin{align} t&= \cos A\cdot \cos B\cdot \cos C = \tfrac14\,(u(v)^2-(v+2)^2) \tag{2}\label{2} . \end{align}

Expressions for the boundary values of $$u$$ are known to be

\begin{align} u_{\min}(v)&= \sqrt{27-(5-v)^2-2\sqrt{(1-2v)^3}} \tag{3}\label{3} ,\\ u_{\max}(v)&= \sqrt{27-(5-v)^2+2\sqrt{(1-2v)^3}} \tag{4}\label{4} . \end{align}

Thus

\begin{align} \min_{\triangle ABC} &(\cos A\cdot \cos B\cdot \cos C) = \tfrac14\,\min_{v\in[0,1/2]} (u_{\min}(v)^2-(v+2)^2) \tag{5}\label{5} \\ &= \tfrac12 \,\min_{v\in[0,1/2]} (v-(1-v)^2-\sqrt{(1-2v)^3}) \tag{6}\label{6} . \end{align}

It can be shown that the function in \eqref{6} is increasing on $$v\in[0,\tfrac12]$$, hence

\begin{align} \tfrac12 \,\min_{v\in[0,1/2]} &(-(1-v)^2+v-\sqrt((1-2v)^3)) \tag{7}\label{7} \\ &= \tfrac12\, (v-(1-v)^2-\sqrt{(1-2v)^3})|_{v=0} =\tfrac12\cdot(-2) =-1 \tag{8}\label{8} . \end{align}

It follows that the minimal value of $$t$$ in \eqref{1} is $$-1$$, which is reached only for degenerate triangles with two zero angles and the third one of $$180^\circ$$. Indeed, \begin{align} \cos0^\circ\cdot\cos0^\circ\cdot\cos180^\circ &=1\cdot1\cdot(-1)=-1 \tag{9}\label{9} . \end{align}