MathLabs

Problem 2

Let a,b,ca,b,c be the side lengths of a triangle whose area is SS. Prove that a2+b2+c2≥4S3a^2+b^2+c^2\ge4S\sqrt{3}. In what case does equality hold?
Step 3 of 6: Square the desired comparison
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4S3=3(4b2c2−(b2+c2−a2)2)4S\sqrt3=\sqrt{3(4b^2c^2-(b^2+c^2-a^2)^2)}
Detailed analysis

Since the area is positive, both sides are nonnegative. Thus it suffices to prove a2+b2+c2a^2+b^2+c^2 is at least the right-hand square root after multiplying the identity by 33.