Inequality Involving Triangle Area and Sides

Date: 05/11/2003 at 21:14:26
From: Bobby
Subject: Area of triangles

Prove that for any triangle with sides a, b, and c and area A,

 a^2+b^2+c^2 greater or equal to (=>) 4sqrt(3)A

I have found for an equilateral triangle that the inequality is equal 
to 4sqrt(3)A.

I want to show that for an isosceles triangle and right triangle the 
inequality is greater than 4sqrt(3)A. In the equilateral case a=b=c 
all fit nicely. Should I use the same system as I found in the 
equilateral case, or should I be using another method?

Date: 05/12/2003 at 09:54:34
From: Doctor Floor
Subject: Re: Area of triangles

Hi, Bobby,

Thanks for your question.

To avoid misunderstanding between angle A and area, I write Area.

From the law of cosines we know:

  -a^2 + b^2 + c^2 = 2bc cosA 
                   = 2bc sinA * cosA/sinA
                   = 4Area * cotA

We may find similar formulas with cotB and cotC. Adding these gives

  a^2 + b^2 + c^2 = 4Area * (cotA + cotB + cotC)

Now let us write 

   S = (a^2+b^2+c^2)/(4Area)

then 

   S = cotA + cotB + cotC 

(in fact, S is the cotangent of Brocard's angle, but that is 
unimportant for the rest of this message).

Now from the following message from the Dr. Math archives:

   Deriving a Trig Identity
   http://mathforum.org/library/drmath/view/54068.html 

we see that in a triangle the identity 

  tanA + tanB + tanC = tanA tanB tanC 

holds, or by division through by tanA tanB tanC

 cotB cotC + cotA cotC + cotA cotB = 1.

By squaring both sides of S = cotA + cotB + cotC  we find 

  S^2 = (cotA)^2 + (cotB)^2 + (cotC)^2 + 2. 

Now we know that

 (cotA - cotB)^2 + (cotB - cotC)^2 + (cotC - cotA)^2 >= 0

and thus

 2((cotA)^2 + (cotB)^2 + (cotC)^2) 
 - 2(cotB cotC + cotA cotC + cotA cotB) >=0

or

 2(S^2 - 2) - 2 >= 0 
 2S^2 - 6 >= 0
 S^2 - 3 >= 0

and we find that S>= sqrt(3) (as it must be positive), which is 
exactly what you were looking for.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/