Prove cotA+cotB+cotC > = square root of 3

Given cotA + cotB + cotC = sqrt3

to prove triangle ABC is equilateral

we prove this by assuming ABC to be equialteral and establishing the truth

of the statement cotA + cotB + cotC = sqrt3

since ABC is equialteral angleA=angleB=angleC=60 degrees

cotA=cotB=CotC = cot60 = 1/sqrt3

therefore cotA + cotB + cotC = 1/sqrt3 + 1/sqrt3 + 1/sqrt3

=3/sqrt3

=sqrt3 which is equal to the RHS (right hand side) of the expression

hence our assumption that ABC is equilateral is true