Let ABC be the three sides of △ ABC, and s be the area of triangle. Prove that C ∧ 2-A ∧ 2-B ∧ 2 + 4AB ≥ (4 √ 3) s

The cosine theorem C ^ 2 = a ^ 2 + B ^ 2-2abcosc is transformed into C ^ 2-A ^ 2-B ^ 2 + 4AB = 4ab-2abcosc triangle area s = 1 / 2absinc. Substituting into the inequality, 4ab-2abcosc ≥ 4 √ 3 × 1 / 2absinc is equivalent to 2-cosc ≥ √ 3sinc, i.e. 1 ≥ √ 3 / 2sinc + 1 / 2cosc, i.e. 1 ≥ sin (c + 30)

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