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 5 of 6: Reduce to a standard quadratic inequality
Previous stepNext step
(A+B+C)2−(6AB+6BC+6CA−3A2−3B2−3C2)=4(A2+B2+C2−AB−BC−CA)(A+B+C)^2-(6AB+6BC+6CA-3A^2-3B^2-3C^2)=4(A^2+B^2+C^2-AB-BC-CA)
Detailed analysis

After squaring and collecting terms, the desired inequality is exactly A2+B2+C2≥AB+BC+CAA^2+B^2+C^2\ge AB+BC+CA.