If a2+b2+c2=1 , then ab+bc+ca lies in the interval

Given that a2+b2+c2=1----(1) we know that (a+b+c)2≥0⇒ a2+b2+c2+2ab+2bc+2ca≥0⇒ 2(ab+bc+ca)≥−1  Using (1)⇒ ab+bc+ca≥−1/2-----(2) Also we know that, 12(a−b)2+(b−c)2+(c−a)2≥0⇒ a2+b2+c2−ab−bc−ca≥0⇒ ab+bc+ca≤1    using(1)⇒ ab+bc+ca≤1----(3) Combining (2) and (3), we get −1/2≤ab+bc+ca≤1⇒ ab+bc+ca∈-12,1