If a2+b2+c2=1, then bc+ca+ab lies in the interval
We have a2+b2+c2+2(bc+ca+ab)=(a+b+c)2≥0
⇒1+2(bc+ca+ab)≥0⇒bc+ca+ab≥−12
Also since A.M. ≥ G.M., for a, b, cwe get
b2+c22≥√b2c2=bc,c2+a22≥√c2a2=ca and
a2+b22≥√a2b2=ab
Add all three inequalities to get
⇒a2+b2+c2≥ab+bc+ca
⇒1≥ab+bc+ca
So, the interval is [−12,1].