Prove this inequality: $|a_1b_1+a_2b_2+\cdots+ a_nb_n|\leq 1$ for two normalised vectors  Help me prove this inequality: $$|a_1b_1+a_2b_2+\cdots+a_nb_n|\leq 1$$
if
$$\begin{align*} a_1^2+a_2^2+\cdots+a_n^2=1, \\
b_1^2+b_2^2+\cdots+b_n^2=1.\end{align*}$$
 A: Note that by the Cauchy-Schwarz inequality, we have:
$$\left(\sum_{i=1}^{n}{a_{i}b_{i}}\right)^{2}\leq\left(\sum_{i=1}^{n}{a_{i}^{2}}\right)\left(\sum_{i=1}^{n}{b_{i}^{2}}\right)$$
Note that if $\sum_{i=1}^{n}{a_{i}^{2}}=\sum_{i=1}^{n}{b_{i}^{2}}=1$, then we have:
$$\left(\sum_{i=1}^{n}{a_{i}b_{i}}\right)^{2}\leq(1\times1=1)$$
This implies (due to the property that $|x|=\sqrt{x^{2}}$):
$$\left|\sum_{i=1}^{n}{a_{i}b_{i}}\right|\leq1$$
Which I assume is what you intended to write.
A: We observe that
$$
(|a_i|-|b_i|)^2\geq 0 \Rightarrow (a_i)^2+(b_i)^2\geq 2|a_ib_i|
$$
for all $i=\overline{1,n}$. Therefore,
\begin{eqnarray*}
\left|\sum_{i=1}^{n}a_ib_i\right|&\leq&\sum_{i=1}^{n}|a_ib_i|\\
&\leq& \sum_{i=1}^{n}\frac{(a_i)^2+(b_i)^2}{2}\\
&=&\frac{\displaystyle\sum_{i=1}^{n}(a_i)^2+\sum_{i=1}^{n}(b_i)^2}{2}=1.
\end{eqnarray*}
The proof is completed.
A: Alternatively, using the definition of the dot product
$a\cdot b = |a||b|\cos\theta$,
where $\theta$ is the angle between the two vectors.
Since $|a|=|b|=1$, then
$a\cdot b=\cos\theta$.
As $|\cos\theta|\leq1$, $|a\cdot b|=|a_1b_1+\ldots +a_nb_n|\leq 1$.
