A sum of triple quadratic averages has an upper bound in the total

Proved

WorkbookSource.base_18112

by wamlart · Sep 13, 2026 · Mathlib 0df444a (Lean v4.33.1)

lean-workbooksource-checked

Prove that for positive numbers a, b, c, and d, the following inequality holds: $\frac{4(b^2+c^2+d^2+a^2)}{c+a+b+d} \geq \frac{b^2+c^2+d^2}{c+d+b}+\frac{c^2+d^2+a^2}{c+d+a}+\frac{d^2+a^2+b^2}{b+d+a}+\frac{a^2+b^2+c^2}{a+b+c}.$

Source: InternLM Lean-Workbook, record lean_workbook_18112 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_18112 (a b c d : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hd : 0 < d) : (4 * (b^2 + c^2 + d^2 + a^2)) / (c + a + b + d) ≥ (b^2 + c^2 + d^2) / (c + d + b) + (c^2 + d^2 + a^2) / (c + d + a) + (d^2 + a^2 + b^2) / (b + d + a) + (a^2 + b^2 + c^2) / (a + b + c)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_18112; Apache-2.0

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