6. Let real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=1$, and let the maximum and minimum values of $a b+b c+c a$ be $M$ and $m$, respectively, then $M-m=$
Answer $\frac{3}{2}$.
Analysis According to the Cauchy-Schwarz inequality $(a b+b c+c a)^{2} \leq\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+a^{2}\right)=1$, so $M=1$, or according to,
$$
2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a)=(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0
$$

we can also get $M=1$, by
$$
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \geq 0
$$

we can get $a b+b c+c a \leq-\frac{1}{2}$, i.e., $m=-\frac{1}{2}$, so $M-m=\frac{3}{2}$