2
$\begingroup$

Suppose $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb R$. Since for all $x,y \in \mathbb R$,
$$(|x|-|y|)^2 \geq 0 \Rightarrow x^2+y^2 \geq 2|xy|$$ we can conclude that $$a_i^2+b_i^2 \geq 2|a_ib_i|$$
for $i=1,2,3$. If $a_1^2+a_2^2+a_3^2=1$ and $b_1^2+b_2^2+b_3^2=1$, then
$\sum_{i=1}^3 a_i^2+b_i^2 \geq \sum_{i=1}^3 2|a_ib_i| \Rightarrow 2 \geq 2(|a_1b_1|+|a_2b_2|+|a_3b_3|)\geq 2(|a_1b_1+a_2b_2+a_3b_3|) \Rightarrow |a_1b_1+a_2b_2+a_3b_3| \leq 1 $

In the final inequality, equality holds iff $|a_i|=|b_i|$ for $i=1,2,3$ and $a_1b_1$,$a_2b_2$ and $a_3b_3$ are all of the same sign. But then my book says that these conditions are equivalent to $a_i=b_i$ for $i=1,2,3$. However, Taking $a_1=b_1=0,a_2=\frac{1}{\sqrt 2}=a_3$ and $b_2=-a_2=b_3$,we get that equality does hold contrary to the conditions of the book.

What did I miss? What is on with this condition $a_i=b_i$ for $i=1,2,3$?

$\endgroup$
1
  • 3
    $\begingroup$ You are right: Equality holds if either $a_i = b_i$ for all $i$, or $a_i = -b_i$ for all $i$. $\endgroup$
    – Martin R
    Commented Oct 20, 2021 at 7:24

1 Answer 1

0
$\begingroup$

You are right: Equality holds if either $a_i = b_i$ for all $i$, or $a_i = -b_i$ for all $i$.

If the $a_ib_i$ have all the same sign and $|a_i|=|b_i|$ for all $i$ then the numbers $$ \{ \frac{a_i}{b_i} \mid b_i \ne 0 \} $$ have the same sign and the modulus one, so they are all equal to $+1$ or all equal to $-1$. It follows that $$ (a1, a_2, a_3) = \pm (b_1, b_2, b_3) \, . $$


One can also use the Cauchy-Schwarz inequality: $\sum_{i=1}^3 a_i^2 = \sum_{i=1}^3 b_i^2 = 1$ implies $$ \sum_{i=1}^3 |a_i b_i| \le \sqrt {\sum_{i=1}^3 a_i^2} \cdot \sqrt {\sum_{i=1}^3 b_i^2} = 1 \cdot 1 = 1 $$ with equality if and only if the vectors $\vec a = (a_1, a_2, a_3)$ and $\vec b = (b_1, b_2, b_3)$ are linearly dependent. Since these vectors have the same length, this is the case if and only if $$ \vec a = \vec b \text{ or } \vec a = - \vec b \, . $$

$\endgroup$

You must log in to answer this question.