Question 92
Let $b_{1}, \ldots, b_{n}$ be nonnegative real numbers such that $b_{1}+b_{2}+\ldots+b_{n}=2$. Let $a_{0}, a_{1}, \ldots, a_{n}$ be real numbers such that $a_{0}=a_{n}=0$. Suppose that for every $i$ between 1 and $n$ inclusive,
$$
\left|a_{i}-a_{i-1}\right| \leqslant b_{i}
$$
Show that
$$
\left(a_{0}+a_{1}\right) b_{1}+\left(a_{1}+a_{2}\right) b_{2}+\ldots+\left(a_{n-1}+a_{n}\right) b_{n} \leqslant 2
$$
$$
\left|a_{i}-a_{i-1}\right| \leqslant b_{i}
$$
Show that
$$
\left(a_{0}+a_{1}\right) b_{1}+\left(a_{1}+a_{2}\right) b_{2}+\ldots+\left(a_{n-1}+a_{n}\right) b_{n} \leqslant 2
$$