<s>[INST] Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$ 
In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$

(note: we can also discover the value of $\cos A\cos B+\cos B\cos C+\cos C\cos A$ using the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$)
Since the last time I've posted this question (the original thread is now deleted), I've reflected a bit on the suggestions of several users. First, I included relevant informations and defintion and second I did try to use the cosine law, but It did not give me help.
I was referred by a friend to the identities
$$\begin{align}
a+b+c &= 2s \tag{1} \\[4pt]
ab+ac+bc &= s^2+r^2+4rR \tag{2} \\[4pt]
abc &= 4Rrs \tag{3}
\end{align}$$ The first and third facts are obvious, while the second I do not know for sure to be true (although it probably is) and appears to model the numerator of the first identity in $\cos A\cos B\cos C$.
Any other idea?
 [/INST] Use the result $1+\frac rR=\cos A+\cos B+\cos C$ in the derivation below
\begin{align}
& \frac{s^2-(2R+r)^2}{4R^2}\\
= &\frac14\left(\frac{a+b+c}{2R}\right)^2-\frac14\left(2+\frac rR\right)^2\\
=& \frac14\left[ (\sin A +\sin B +\sin C)^2 -(1+ \cos A +\cos B +\cos C)^2 \right]\\
=& \frac14\left[(\sin^2A-\cos^2 A)+(\sin^2 B-\cos^2 B) + (\sin^2 C-\cos^2 C)\right.\\
& \left.-1 -2( \cos A +\cos B +\cos C)-2(\cos (A+B)+\cos (B+C) +\cos (C+A)) \right]\\
= &\frac14( -\cos2A - \cos2B -\cos2C-1)\\
= &\frac14( -2\cos(A+B)\cos(A-B) -2\cos^2C +1 -1)\\ 
= &\frac12\cos C(\cos(A-B)+\cos(A+B))\\ 
= &\>\cos A \cos B\cos C\\ 
\end{align}
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