1. Given $a, b, c$ are real numbers, and $2a + b + c = 5, b - c = 1$. Then the maximum value of $ab + bc + ca$ is
From $2 a+b+c=5, b-c=1$, we can obtain
$$
b=3-a, c=2-a \text {. }
$$

Therefore, $a b+b c+c a$
$$
\begin{array}{l}
=a(3-a)+(3-a)(2-a)+a(2-a) \\
=-a^{2}+6 \leqslant 6 .
\end{array}
$$