**·**2011-08-14 23:49:42

After 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 Î£cosA=2-(Î£1cosA)<-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)â‰¥12.

An elementary soln. is very welcome. [1]