Let s,R and r be the semi-perimeter, Circumradius and Inradius of triangle ABC respectively. Then show that:

1) 8r²(cos(A-B)+cos(B-C)+cos(C-A)) ≤ s²-3r²

2) (1/(sinA*sinB))+(1/(sinB*sinC))+(1/(sinC*sinA)) ≥ 4

Hi!

I will give a solution for the second inequalitie!

The relation is the same with

sin(A)+sin(B)+sin(C) >= 4*sin(A)*sin(B)*sin(C)

But in a triangle  (after some calculus)

sin(A)+sin(B)+sin(C) = 4*cos (A/2)*cos(B/2)*cos(C/2)

So, the puzzle can be transform in :

sin(A/2)*sin(B/2)*sin(C/2) <= 1/8

To prove this i use the relations :

sin(A/2) = sqrt((p-b)*(p-c)/(b*c)), sin(B/2) = sqrt((p-a)*(p-c)/(a*c)), sin(C/2) = sqrt((p-a)*(p-b)/(a*b))

So i found

8*(p-a)*(p-b)*(p-c) <= a*b*c  or

(-a+b+c)*(a-b+c)*(a+b-c) <= a*b*c

If i note

-a+b+c = x,a-b+c = y,a+b-c = z with x,y,z > 0

the relation can be write like this :

8*x*y*z <= (x+y)*(y+z)*(z+x)

This relation is well know and can be prove using

x+y>=2*sqrt(x*y), y+z>=2*sqrt(y*z), z+x>=2*sqrt(z*x)

and multiply toghether!

Have a nice day!

 

 

 

 

 

Posted by Chesca Ciprian on 2008-06-15 03:58:10