This problem was asked in IMO 1961, Hungary.

Q: Let a,b,c be the lengths of a triangle ABC whose area is S. Prove that a^2+b^2+c^2 \geq 4S \sqrt{3}. In what case does equality hold?

Solution: We can express the area in terms of the sides and angles of the triangle:

S=\frac{1}{2}bc sin A

Also, the cosine rule says that

a^2=b^2+c^2-2bc cos A

Therefore,

a^2+b^2+c^2 \geq 4S \sqrt{3} \iff b^2+c^2 \geq bc( \sqrt3 sin A + cos A) \\ \iff (b-c)^2+2bc(cos (A-60^0) \geq 0

Note that equality will hold above if b=c and a=60^0, that is, when the triangle is equilateral.

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