Trigonometry (For what values of $(\cos A-1/5)(\cos B-1/5)(\cos C-1/5) \leq k$ hold in all triangles $ABC$? For what  values  of  $k$ does
$$
\bigl(\cos A-\frac{1}{5}\bigr)\bigl(\cos B-\frac{1}{5}\bigr)\bigl(\cos C-\frac{1}{5}\bigr ) \leqslant k\:
$$
hold  in  all triangles $ABC$?
 A: \begin{align}
(\cos A-\tfrac15)(\cos B-\tfrac15)(\cos C-\tfrac15)
\le k
\tag{1}\label{1}
.
\end{align}
Given that for $\triangle ABC$ we know its semiperimeter $\rho$,
inradius $r$ and circumradius $R$,
let $u=\rho/R$, $v=r/R$. It is known that
for a valid triangle (including degenerate cases)
\begin{align}
v&\in[0,\tfrac12]
,\\
u&\in[u_{\min}(v),u_{\max}(v)]
\tag{2}\label{2}
,
\end{align}
where the boundary expressions
\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}
correspond to two kinds of isosceles triangles with given $v$.
Consider the left-hand side of expression \eqref{1}
in terms of $u,v$;
\begin{align} 
(\cos A-\tfrac15)
&(\cos B-\tfrac15)(\cos C-\tfrac15)
=(\cos A\cos B\cos C)
\tag{5}\label{5}
\\
&-\tfrac15(\cos A\cos B+\cos B\cos C+\cos C\cos A)
\\
&+
\tfrac1{25}(\cos A+\cos B+\cos C)
-\tfrac1{125}
\tag{6}\label{6}
,\\
&=
(\tfrac14\,(u^2-(v+2)^2))
-\tfrac15(\tfrac14\,(u^2+v^2)-1)
\\
&\phantom{=}
+\tfrac1{25}(v+1)
-\tfrac1{125}
\tag{7}\label{7}
,\\
&=
\tfrac15\,u^2-\tfrac3{10}\,v^2-\tfrac{24}{25}\,v-\tfrac{96}{125}
\tag{8}\label{8}
,
\end{align}
which leads to
\begin{align}
\max_{v\in[0,\frac12]}&
(\tfrac15\,u(v)^2-\tfrac3{10}\,v^2-\tfrac{24}{25}\,v-\tfrac{96}{125})
\tag{9}\label{9}
\\
&=\max_{v\in[0,\frac12]}
(\tfrac15\,u_{\max}(v)^2-\tfrac3{10}\,v^2-\tfrac{24}{25}\,v-\tfrac{96}{125})
\tag{10}\label{10}
\\
&=(\tfrac15\,u_{\max}(v)^2-\tfrac3{10}\,v^2-\tfrac{24}{25}\,v-\tfrac{96}{125})
\Big|_{v=0}
\tag{11}\label{11}
\\
&=\tfrac{4}{125}
.
\end{align}
Similarly, a minimal value of the left-hand side of \eqref{1}
can be found to be $-\tfrac{96}{125}$.
