α and β are positive real numbers such that 1/α + 1/β = 1. Prove that the line mx + ny = 1 with m, n positive reals is tangent to the curve xα + yα = 1 in the first quadrant (x, y ≥ 0) iff mβ + nβ = 1.
Suppose mx + ny = 1 is tangent to the curve. Suppose it touches at (a, b). Differentiating, we see that the tangent at (a, b) is aα-1x + bβ-1 = 1, so m = aα-1, n = bβ-1. Hence, using αβ - β = α, we have that mβ + nβ = aα + bα = 1.
Conversely, suppose that mβ + nβ = 1. Take a = mβ/α, b = nβ/α. Then aα + bα = 1, so (a, b) lies on the curve in the first quadrant. Its tangent is Mx + Ny = 1, where M = aα-1, N = bβ-1. But a = mβ/α and β/α (α - 1)= 1, so M = m. Similarly, N = n. Thus we have established that mx + ny = 1 is tangent to the curve in the first quadrant as required.
26th Putnam 1965
© John Scholes
25 Jan 2002