But how ?
Im stuck at this point
Any time you have a theorem that involves proving something for every integer greater than some base integer, you should consider a proof by weak mathematical induction.
There are two points about such proofs. One is that you can never be sure whether they are going to work until you try. So if you do not see any other proof, start working on the assumption that induction will work. Don't just stare at the thing; do the work, starting at n = 1.
Second, virtually all proofs by induction work similarly. After you have proved it for 1, you can assume that it is true for at least one positive integer. Let k be ANY such integer. So the theorem is true for k. Now set up the pieces for k + 1 and REDUCE them to the desired expressions in k and an expression in k + 1. When you are dealing with complex expressions, it frequently helps to replace them with simple variables to make the algebra more manageable. So here it is how it is done.
\(\displaystyle \displaystyle n = 1 \implies \left ( \sum_{j=1}^n \{a_jb_j\}^2 \right ) = (a_1b_1)^2 = a_1^2b_1^2.\)
\(\displaystyle \displaystyle n = 1 \implies \left ( \sum_{j=1}^n a_j^2 \right ) \left ( \sum_{j=1}^n b_j^2 \right ) = a_1^2b_1^2.\)
\(\displaystyle \displaystyle \therefore n = 1 \implies \left ( \sum_{j=1}^n \{a_jb_j\}^2 \right ) \le \left ( \sum_{j=1}^n a_j^2 \right ) \left ( \sum_{j=1}^n b_j^2 \right ).\)
\(\displaystyle \text {THUS, } \exists\ k \in \mathbb Z \text { such that } k \ge 1\\
\text { and } \displaystyle \left ( \sum_{j=1}^k \{a_jb_j\}^2 \right ) \le \left ( \sum_{j=1}^k a_j^2 \right ) \left ( \sum_{j=1}^k b_j^2 \right ).\)
\(\displaystyle \displaystyle \text {Let } c = \left ( \sum_{j=1}^k \{a_jb_j\}^2 \right ),\ d = \left ( \sum_{j=1}^k a_j^2 \right ),\ \text {and } e = \left ( \sum_{j=1}^k b_j^2 \right ).\)
\(\displaystyle \text {Note that } 0 \le c \le de.\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^{k+1} \{a_jb_j\}^2 \right ) = \left ( \sum_{j=1}^k \{a_jb_j\}^2 \right ) + a_{k+1}^2b_{k+1}^2 = c + a_{k+1}^2b_{k+1}^2 .\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^{k+1} a_j^2 \right ) \left ( \sum_{j=1}^{k+1} b_j^2 \right ) = \left \{ \left ( \sum_{j=1}^k a_j^2 \right ) + a_{k+1}^2 \right \}\left \{\left ( \sum_{j=1}^k b_j^2 \right ) + b_{k+1}^2 \right \} = \)
\(\displaystyle (d + a_{k+1}^2)(e + b_{k+1}^2) = de + a_{k+1}^2e + b_{k+1}^2d + a_{k+1}^2b_{k+1}^2 \implies\)
\(\displaystyle \displaystyle de + a_{k+1}^2b_{k+1}^2 = \left ( \sum_{j=1}^{k+1} a_j^2 \right ) \left ( \sum_{j=1}^{k+1} b_j^2 \right ) - (a_{k+1}^2e + b_{k+1}^2d).\)
\(\displaystyle \text {But } 0 \le (a_{k+1}^2e + b_{k+1}^2d)\ \because \text { every term is non-negative.}\)
\(\displaystyle \displaystyle \therefore de + a_{k+1}^2b_{k+1}^2 \le \left ( \sum_{j=1}^{k+1} a_j^2 \right ) \left ( \sum_{j=1}^{k+1} b_j^2 \right ).\)
\(\displaystyle \displaystyle \therefore c + a_{k+1}^2b_{k+1}^2 \le \left ( \sum_{j=1}^{k+1} a_j^2 \right ) \left ( \sum_{j=1}^{k+1} b_j^2 \right ) \ \because \ c \le de.\)
\(\displaystyle \displaystyle \text {But } \left ( \sum_{j=1}^{k+1} \{a_jb_j\}^2 \right ) = c + a_{k+1}^2b_{k+1}^2 .\)
\(\displaystyle \displaystyle \therefore \left ( \sum_{j=1}^{k+1} \{a_jb_j\}^2 \right ) \le \left ( \sum_{j=1}^{k+1} a_j^2 \right ) \left ( \sum_{j=1}^{k+1} b_j^2 \right ).\)
\(\displaystyle \displaystyle \text {THUS, by induction, for all } n \ge 1,\ \left ( \sum_{j=1}^n \{a_jb_j\}^2 \right ) \le \left ( \sum_{j=1}^n a_j^2 \right ) \left ( \sum_{j=1}^n b_j^2 \right ).\)
