ABC triangle ABC has three sides, s is the area of triangle, prove C & sup2; - A & sup2; - B & sup2; + 4AB ≥ 4 √ 3S
By cosine theorem: C ^ 2-A ^ 2-B ^ 2 = - 2abcosc
S=(1/2)absinC
The inequality becomes: - 2abcosc + 4AB > = 2 √ 3absinc
It is proved that 2 √ 3absinc + 2abcosc
RELATED INFORMATIONS
- 1. Let a, B, C 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
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