I am learning about Einstein (index) notation, and I am trying to write Cauchy-Schwartz inequality as such. If we have $(\boldsymbol{a} \cdot \boldsymbol{b})^2 \le ||\boldsymbol{a}||^2 \ ||\boldsymbol{b}||^2$, and knowing that whenever an index is repeated in a term, a summation over this index is implied, we can write: $$(\boldsymbol{a} \cdot \boldsymbol{b})^2 = (a_1b_1 + a_2b_2 + a_3b_3)^2 = \sum_{i=1}^3 a_ib_i = a_i b_i $$ $$ ||\boldsymbol{a}||^2 = a_1a_1 + a_2a_2 + a_3a_3 = a_ia_i$$ Equivalently $||\boldsymbol{b}||^2 = b_i b_i$
So, I thought that the answer would be $(a_ib_i)^2 \le (a_ia_i) (b_ib_i)$, yet my textbook gives the answer as $(a_ib_i)^2 \le \boxed{a_j ^2 b_k ^2}$
Can someone explain how can we get to the right hand side of the inequality? (or point out if I have a misunderstanding for the index notation).
With index notation you have to be careful about repeating indices. First as @ClementC pointed out $$ \|\mathbf{a}\| = a_1^2 + a_2^2 + a_3^2 = a_ja_j = a_j^2$$ where we understand that even though $a_i^2$ only has one explicit index since it is squared we can write $a_j^2 = a_ja_j$ and so still sum over $j$. and similarly $\|\mathbf{b}\| = b_k^2$. Here we choose a different index to avoid issues later. The Cauchy-Schwartz inequality is then, as stated in your textbook, $$ (a_ib_i)^2 \le a_j^2b_k^2. $$ The fact that $a$ and $b$ have different indices is important. For example the sum on the right includes terms like $$ a_1a_1b_2b_2 $$ whereas if we instead had $a_i^2b_i^2$ then $a$ and $b$ always have the same index so this term would not be included. In general if you have multiple terms with indices and then combine them into some larger term you should make sure not to have the same indices in each term unless you explicitly want no cross terms.