# Prove $|a_1b_1+a_2b_2+\cdot\cdot\cdot + a_nb_n|\leq 1$ [duplicate]

Prove this inequality: $|a_1b_1+a_2b_2+\cdot\cdot\cdot + a_nb_n|\leq 1$

if

$a_1^2+a_2^2+\cdot\cdot\cdot+a_n^2=1$

$b_1^2+b_2^2+\cdot\cdot\cdot+b_n^2=1$

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## marked as duplicate by Qiaochu YuanSep 9 '12 at 19:05

Obviously there is some mistake or a typo, since this is not true. For $a_1=b_1=1$ (and all remaining equal to zero) you have $|a_1b_1+\dots+a_nb_n|=1$. And absolute value $|x|$ can be $\le0$ only if $x=0$. –  Martin Sleziak Sep 9 '12 at 18:54
Well, no. ${}{}{}{}$ –  Will Jagy Sep 9 '12 at 18:55

Before we prove this inequality first prove the following inequality: $|ab|\geq\frac{a^2+b^2}{2}$.

Given by the inequality we have:

$2|ab|\leq a^2+b^2$

$2|a||b|\leq a^2+b^2$ $a^2+b^2-2|a||b|\geq 0$

or

$(a-b)^2\geq 0$

is true for all $a$ and $b$. Now go back and the prove given inequalty:

$|a_1b_1+a_2b_2+\cdot\cdot\cdot + a_nb_n|\leq |a_1b_1|+|a_2b_2|+\cdot\cdot\cdot+|a_nb_n|\leq|a_1||b_1|+|a_2||b_2|+\cdot\cdot\cdot+|a_n||b_n|\leq \frac{a_1^2+b_1^2}{2}+\frac{a_2^2+b_2^2}{2}+\cdot\cdot\cdot+\frac{a_n^2+b_n^2}{2}$=$\frac{a_1^2+b_1^2+a_2^2+b_2^2\cdot\cdot\cdot+a_n^2+b_n^2}{2}$=$\frac{(a_1^2+a_2^2+\cdot\cdot\cdot+a_n^2)+(b_1^2+b_2^2+\cdot\cdot\cdot+b_n^2)}{2}$$\leq \frac{1+1}{2}=1 - Note that by the Cauchy-Schwarz inequality, we have:$$\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 is what I assumed you intended to write.

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