An inequality for triangles

By OldKam on Thursday, December 05, 2002 - 02:46 pm:

Let a, b, c be the three sides opposite to the three angles A, B, C (radian) of a triangle. Prove that p/3 £ (a A+b B+c C)/(a+b+c) < p/2
Oldkam

By Demetres Christofides on Thursday, December 05, 2002 - 05:15 pm:

For the second inequality you need only to use some elementary triangle inequalities. I'll let you find it.

For the first inequality: This is equivalent to proving that a A+b B+c C ł ap/3+bp/3+cp/3.

The first thing to notice is that A+B+C=p.

The second is that a ł b ł c implies that A ł B ł C.

Try to show that if a ł b ł c and x ł y ł z with x+y+z=0 then a x+b y+c z ł 0.

Demetres

By OldKam on Friday, December 06, 2002 - 11:44 am:

I don't understand the last sentence. Where does "x+y+z=0" come from ?

Thanks

Oldkam

By Demetres Christofides on Friday, December 06, 2002 - 01:19 pm:

You can then take x=A-p/3, y=B-p/3 and z=C-p/3.

Demetres

By OldKam on Sunday, December 08, 2002 - 11:48 am:

I know it must be a very easy question though i find it hard to answer. (I'm rather weak in inequality) I still can't show a x+b y+c z ł 0. I've tried to make use of A+B+C=p and x+y+z=0 but can't see where I can use it. Any help or solution would be appreciated.

Thanks

Oldkam

By Demetres Christofides on Sunday, December 08, 2002 - 09:00 pm:

Since x+y+z=0 then z=-x-y. So a x+b y+c z=(a-c)x+(b-c)y.

Since we are assuming that a ł b ł c and x ł y ł z, x+y+z=0 then (a-c) ł 0, (b-c) ł 0 and x+y ł 0.

Demetres

By K.L. Kam on Thursday, December 19, 2002 - 11:30 pm:

Thanks Demetres, got the idea now.

oldkam