2022/077) If $a^2+b^2+c^2 = 1$ find the range of $ab + bc+ca$

We need to find the mininal and maximal of $ab+bc+ca$

we have $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ bc+ ca)$

so $ a^2 + b^2 + c^2 + 2(ab+ bc+ ca) > = 0$

puttig $a^2+b^2+c^2 = 1$ we get

$ 1 + 2(ab+ bc+ ca) > = 0$

or $(ab+bc+ca) >=  - \frac{1}{2}$

Further to find tthe maximum we have $(a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 +b^2+c^2 - ab - bc - ca)$

or $2(a^2 +b^2+c^2 - ab - bc - ca) >= 0$

or $ab+bc+ca <= a^2+b^ + c^2$

or $ab+bc+ca <= 1$

so we have $ab+bc + ca \in [-.5 .. 1]$