If $\lvert a_i\rvert < 1$, $\lambda_i\geq 0$ for $i = 1,\ldots,n$ and $\lambda_1 + \lambda_2 + \cdots + \lambda_n = 1$, show that $$ \lvert\lambda_1a_1 + \lambda_2a_2 + \cdots + \lambda_na_n\rvert < 1. $$
Since $\sum_{i = 1}^n\lambda_i = 1$ and $\lambda_i\geq 0$, we have that $0\leq \lambda_i < 1$. By Cauchy's inequality, $$ \lvert\lambda_1a_1 + \lambda_2a_2 + \cdots + \lambda_na_n\rvert^2 \leq\bigl(\lvert a_1\rvert^2 + \cdots + \lvert a_n\rvert^2\bigr) \bigl(\lvert\lambda_1\rvert^2 + \cdots + \lvert\lambda_n\rvert^2\bigr) $$ Since $\lambda_i^2 < \lambda_i < 1$, we have $\lvert\lambda_1\rvert^2 + \cdots + \lvert\lambda_n\rvert^2 < 1$ and $$ < \lvert a_1\rvert^2 + \cdots + \lvert a_n\rvert^2\tag{1} $$ How do I show equation $(1)$ is less than $1$?