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