# A problem in understanding the conditions for equality of an inequality

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 $$i=1,2,3$$?

• You are right: Equality holds if either $a_i = b_i$ for all $i$, or $a_i = -b_i$ for all $i$. Commented Oct 20, 2021 at 7:24

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 $$\vec a = \vec b \text{ or } \vec a = - \vec b \, .$$