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 4 of 6: Set the squared side variables
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A=a2, B=b2, C=c2⟹A+B+C≥6AB+6BC+6CA−3A2−3B2−3C2A=a^2,\ B=b^2,\ C=c^2\quad\Longrightarrow\quad A+B+C\ge\sqrt{6AB+6BC+6CA-3A^2-3B^2-3C^2}
Detailed analysis

Substituting A=a2,B=b2,C=c2A=a^2,B=b^2,C=c^2 and expanding the square root reduces the claim to the displayed inequality. Both sides are nonnegative, so squaring is legitimate.