A triangle has sides a, b, c. The radius of the inscribed circle is r and s = (a + b + c)/2. Show that 1/(s - a)2 + 1/(s - b)2 + 1/(s - c)2 ≥ 1/r2.
Let A = s-a, B = s-b, C = s-c. Then (A-B)2 ≥ 0 with equality iff a = b. Hence 2/(AB) ≤ 1/A2 + 1/B2. Similarly for 2/(BC) and 2/(CA). Hence 1/(BC) + 1/(CA) + 1/(AB) ≤ 1/A2 + 1/B2 + 1/C2 with equality iff the triangle is equilateral.
Now 1/(BC) + 1/(CA) + 1/(AB) = (A + B + C)/(ABC). But A + B + C = s. By Heron's theorem, sABC = k2, where k is the area of the triangle. Also (dividing the triangle into three by connecting the incentre to each vertex, and considering the area of each part) k = rs. Hence s/(ABC) = 1/r2, giving the required result.
27th Putnam 1966
© John Scholes
25 Jan 2002