Let a, b, c be the sides of a triangle and A its area. Prove that:
a2 + b2 + c2 ≥ 4√3 A
When do we have equality?
Solution
One approach is a routine slog from Heron's formula. The inequality is quickly shown to be equivalent to a2b2 + b2c2 + c2a2 ≤ a4 + b4 + c4, which is true since a2b2 ≤ (a4 + b4)/2. We get equality iff the triangle is equilateral.
Another approach is to take an altitude lying inside the triangle. If it has length h and divides the base into lengths r and s, then we quickly find that the inequality is equivalent to (h - (r + s)√3/2)2 + (r - s)2 ≥ 0, which is true. We have equality iff r = s and h = (r + s)√3/2, which means the triangle is equilateral.
A third solution is due to Jonathan Mizrahi (somewhat adapted):
We have b2 + c2 >= 2bc with equality iff b = c. Also for any angle x in the range 0o to 180o we have 2bc ≥ 2bc sin(X + 30o) with equality iff X = 60o. So taking X to be the angle between the sides b and c (we cannot call it A because A is already used to mean the area in this question!) we have that b2 + c2 ≥ bc sin(X + 30o) with equality iff the triangle is equilateral. Now 2 sin(X + 30o) = √3 sin X + cos X, so using the cosine rule a2 = b2 + c2 - 2bc cos X, we get the required inequality.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
19 Sep 1998
Last corrected/updated 24 Sep 2003