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 6 of 6: Finish by pairwise AM–GM and identify equality
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A2+B22≥AB,B2+C22≥BC,C2+A22≥CA\frac{A^2+B^2}{2}\ge AB,\quad\frac{B^2+C^2}{2}\ge BC,\quad\frac{C^2+A^2}{2}\ge CA
Detailed analysis

Adding the three AM–GM inequalities gives A2+B2+C2≥AB+BC+CAA^2+B^2+C^2\ge AB+BC+CA. Equality requires A=B=CA=B=C, hence a=b=ca=b=c; the triangle is equilateral, and direct substitution shows equality holds.