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In any triangle $ABC$

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

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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}

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