# 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*}

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Is it a pen slip of $|a_1b_1+\cdots+a_nb_n|\le1$ by Cauchy-Schwarz inequality? – Frank Science Jul 15 '12 at 14:58

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 $|\cos\theta|\leq1$, $|a\cdot b|=|a_1b_1+\ldots +a_nb_n|\leq 1$.

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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.

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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.

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