Let for a differentiable function f:(0, ∞) → R, f(x)-f(y) ≥ log ₑ(x/y)+x-y, ∀ x, y ∈(0, ∞). Then Σₙ₌₁²⁰ f′(1/n²) is equal to ____________.

Solution

Maths sir · 4 steps

Move everything to one side and you get a function that can never change, so it must be constant.

1. Build a new function. Since logₑ x/y = ln x - ln y, the inequality becomes f(x) - ln x - x ≥ f(y) - ln y - y. Call g(t) = f(t) - ln t - t, so g(x) ≥ g(y) for every x, y > 0.
2. Swap x and y. The condition holds for all pairs, so it also gives g(y) ≥ g(x). Both together force g(x) = g(y), so g is a constant c.

   f(x) = ln x + x + c ⇒ f'(x) = 1/x + 1

3. Put in the points. At x = 1/n² we get f'(1/n²) = n² + 1. Summing from 1 to 20 gives Σ n² + 20 = (20 × 21 × 41)/6 + 20 = 2870 + 20.
4. So. The sum is 2890.

The idea to keep. If an inequality holds for every pair x, y, swap them. When the swapped version points the other way, the inequality is really an equality and the function is pinned down.

Answer from the NTA final answer key. Solution by PrepWiser.