If a, b, c are real numbers $$a^{2} + b^{2} + c^{2} = 1$$, then the set of values $$ab+bc+ca$$ can take is:

We know, 

$$\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)$$We know, 

so, $$\left(a+b+c\right)^2=1+2\left(ab+bc+ac\right)$$

or, $$ab+bc+ac=\dfrac{\left(a+b+c\right)^2-1}{2}$$

Now, the least value of a square can be 0

so, the least value of $$ab+bc+ac=\dfrac{0-1}{2}=-\dfrac{1}{2}$$

Also, we know, $$A.M.\ge G.M.$$

so, $$\dfrac{a^2+b^2}{2}\ge ab$$

or, $$a^2+b^2\ge2ab$$

similarly, $$b^2+c^2\ge2bc$$

similarly, $$a^2+c^2\ge2ac$$

Adding the above three inequalities we get, $$2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)$$

or, $$a^2+b^2+c^2\ge ab+bc+ac$$

or, $$1\ge ab+bc+ac$$

So, the maximum value of $$ab+bc+ac$$ is 1

so, $$ab+bc+ac\in\ \left[-\dfrac{1}{2},1\right]$$