Question 118
The real numbers $a_i, b_i$, $1 \le i \le n$, are such that
$$
\sum_{i=1}^{n} a_{i}^{2}=1, \sum_{i=1}^{n} b_{i}^{2}=1 \text{ and } \sum_{i=1}^{n} a_{i} b_{i}=0.
$$
Prove that
$$
\left(\sum_{i=1}^{n} a_{i}\right)^{2}+\left(\sum_{i=1}^{n} b_{i}\right)^{2} \leq n.
$$
$$
\sum_{i=1}^{n} a_{i}^{2}=1, \sum_{i=1}^{n} b_{i}^{2}=1 \text{ and } \sum_{i=1}^{n} a_{i} b_{i}=0.
$$
Prove that
$$
\left(\sum_{i=1}^{n} a_{i}\right)^{2}+\left(\sum_{i=1}^{n} b_{i}\right)^{2} \leq n.
$$