According to Heath, this proof is really due to Archimedes. It should be mentioned that it is of course a lot easier to prove the result using trigonometry! The area of a triangle is 1/2ab.sinC, and using c2=a2+b2-2ab.cosC, and sin2+cos2=1, the result follows in a few lines.


Here they are:

Note that the ab outside cancels that inside, so all we have to do is rearrange the numerator inside. For convenience, I change the overall sign of the expression to get:


(a2 + b2 c2)2 - 4a2b2 = (a2-b2)2 2c2(a2 +b2) + c4


= ((a + b)2 c2)((a b)2 c2)

= (a + b + c)(a + b c)(a b + c)(a b c)