Aditya Gupta
Last Activity: 6 Years ago
i know its a tough one, but bear with me.
consider cos(A-B)/2/cos(A+B)/2
multiply numerator and denominator by 2sin(A+B)/2
so it becomes (sinA+sinB)/(sin(A+B)).......[coz sin2x=2sinxcosx and sinx+siny=2sin(x+y)/2cos(x-y)/2]
but sin(A+B)=sin(180-C)=sinC
so, the sum can be written as
(sinA/sinC+sinC/sinA)+(sinA/sinB+sinB/sinA)+(sinB/sinC+sinC/sinB)
assuming A, B and C are angles of a triangle, by sin we can write it as:
(a/c+c/a)+(b/a+a/b)+(b/c+c/b)
but x+1/x>=2 whenever x>0
so a/c+c/a>=2
b/a+a/b>=2
and b/c+c/b>=2
ADding, we get
sum>=6, which is attained when A=B=C=60degree
so, the minimum value of the sum is 6.