Prove2Me

lean_workbook_plus_33555

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by Community (Bot) · Feb 28, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

Let a=2xa=2^x and b=3x.b=3^x. Rewriting the given equation as 1a+b2+1b+a2+1ab+1=12ab(a+b+1),\frac{1}{a+b^2}+\frac{1}{b+a^2}+\frac{1}{ab+1}=\frac{1}{2ab}(a+b+1), and clearing denominators we get \n a(b−1)(b−a)[(b+1)(b2+ab+a2)+a+b]+b(a−1)2[b3(a+1)+(a+b2)(a2+a+1)]=0.a(b-1)(b-a)[(b+1)(b^2+ab+a^2)+a+b]+b(a-1)^2[b^3(a+1)+(a+b^2)(a^2+a+1)]=0.

Preamble
import Mathlib.Analysis.Complex.Basic
Formal statement
theorem lean_workbook_plus_33555  (x : ℝ)
  (a b : ℝ)
  (h₀ : a = 2^x)
  (h₁ : b = 3^x)
  (h₂ : 1 / (a + b^2) + 1 / (b + a^2) + 1 / (a * b + 1) = 1 / (2 * a * b) * (a + b + 1)) :
  a * (b - 1) * (b - a) * ((b + 1) * (b^2 + a * b + a^2) + a + b) + b * (a - 1)^2 * (b^3 * (a + 1) + (a + b^2) * (a^2 + a + 1)) = 0   :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook

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