11After days of waiting for an elementary solution to turn up, I finally decided to try something out-of-syllabus sort.
The point is that certain trigonometric expressions in cosA, cosB, cosC can be optimised if
the given expression can be written down as H(s,p,q)
where s=\sum {\cos A}
p=\sum {\cos A \cos B}
q=\sum {\cos A \cos B \cos C }
Needless to mention A,B,C are angles of a triangle.
Now given this I'll show how exactly the optimisation is done.
First, we will find the interior critical points of H by solving the following system of equations:
\frac{\partial H}{\partial s}+\frac{\partial H}{\partial p}s=\frac{\partial H}{\partial q}+\frac{\partial H}{\partial p}=0
s^2=1+2p-2q
\Delta =-27q^2+18spq+p^2s^2-4s^3q-4p^3>0..............(I)
Usually no such critical point exists.
Next we optimise H on degenerate triangles, i.e s=1, p=q, q\in [-1,0]
...............(II)
Finally optimise H over isosceles triangles - .........(III)
s=2t+1-2t^2
p=t^2+2t(1-2t^2)
q=t^2(1-2t^2) : t \in [0,1]
Now applying AM-GM on the given exp. we have to analyse
\prod{\left(\frac{1+\cos A \cos B}{1+ \cos C} \right)}
Observe that this is infact
H(s,p,q)=\frac{1+p+qs+q^2}{1+p+s+q}
Applying the first algorithm to find the interior critical points, we have
s=2-\frac{p}{q}
Which is not possible since ΣcosA=2-(Σ1cosA)<-1 is a contradiction.
Region over Degenerate triangles does not yield any critical points.
And the region over isosceles triangles gives something like P(t)Q(t)-S(t)X(t)
where P\left(\frac{1}{2} \right)=X\left(\frac{1}{2} \right)=0
Clearly t=12 is a critical point where the expression gets optimised. We're yet to decide whether it's a maximum or minimum. For that we just test the given expression with A=90deg. B=45deg. C=45deg. to arrive at the conclusion that the given exp. is infact minimised at t=12.
Now obtain the values of s,p,q from (III)
Putting the respective values of s,p,q in the exp, we arrive that H(s,p,q)≥12.
An elementary soln. is very welcome. [1]
