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

π / 3 \leq (a A + b B + c C) / (a + b + c) < π / 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 $a A + b B + c C \geq a π / 3 + b π / 3 + c π / 3$ .

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

The second is that $a \geq b \geq c$ implies that $A \geq B \geq C$ .

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

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 - π / 3

y = B - π / 3

and

z = C - π / 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 \geq 0

. I've tried to make use of

A + B + C = π

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 $a \geq b \geq c$ and $x \geq y \geq z$ , $x + y + z = 0$ then $(a - c) \geq 0$ , $(b - c) \geq 0$ and $x + y \geq 0$ .

Demetres

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

Thanks Demetres, got the idea now.

oldkam