### 21st Putnam 1960

Problem A3

Let a, b, g, d, e be arbitary reals. Show that (1 - α) eα + (1 - β) eα+β + (1 - γ) eα+β+γ + (1 - δ) eα+β+γ+δ + (1 - ε) eα+β+γ+δ+ε ≤ k4, where k1 = e, k2 = k1e, k3 = k2e, k4 = k3e (so k4 is 10k with k approx 1.66 million).

Solution

Differentiating shows that the maximum of (1 - x)ex is 1 and occurs at x = 0. Now consider (1 - α) eα + (1 - β) eα + β = eα ( 1 - α + (1 - β) eβ). For any given α, the maximum over β is 1, so we have to maximise eα(2 - α). Differentiating shows that this has maximum k1 = e at α = 1.

Now consider (1 - α) eα + (1 - β) eα + β + (1 - γ) eα + β + γ = eα ( 1 - α + (1 - β) eβ + (1 - γ) eβ + γ). We have just shown that the maximum of (1 - β) eβ + (1 - γ) eβ + γ is k1 at β = 1, γ = 0. So we have to maximise (1 - α + k1) eα. Differentiating shows that this is k2 at α = k1.

Similarly, the maximum of (1 - α) eα + (1 - β) eα + β + (1 - γ) eα + β + γ + (1 - δ) eα + β + γ + δ is k3 at α = k2, β = k1, γ = 1, δ = 0. Finally, the maximum of the desired expression is k4 at α = k3, β = k2, γ = k1, δ = 1, ε = 0.