Are these two forms of Cauchy-Schwarz inequality equivalent? I have Cauchy-Schwarz inequality in the form 
$$\lvert x_1y_1+\dots+x_ny_n\rvert \leq \sqrt{x_1^2+...+x_n^2}\sqrt{y_1^2+...+y_n^2}.\tag{1}$$
But I have also seen the following form:
$$\lvert x_1y_1\rvert +...+ \lvert x_ny_n\rvert \leq \sqrt{x_1^2+...+x_n^2}\sqrt{y_1^2+...+y_n^2}\tag{2}$$
But I don't see why these are equivalent. Isn't it true that in general $$\lvert x_1y_1+\dots+x_ny_n\rvert \neq \lvert x_1y_1\rvert +...+ \lvert x_ny_n\rvert?$$
 A: One can see by triangle inequality that the (2) implies (1). 
Now suppose (1) is true. In particular, it is true for any $(y_1,\cdots,y_n)$. Thus, one can change the sign of $y_i$, so that
$$
|x_1y_1+\cdots+x_ny_n|=|x_1y_1|+\cdots+|x_ny_n|
$$
Since the right hand side of (1) stays the same, one has (2). 
The equivalency between (1) and (2) does not imply that in general, 
$$
|x_1y_1+\cdots+x_ny_n|=|x_1y_1|+\cdots+|x_ny_n|.
$$
and thus there is no contradiction.

To see what I mean by "change the sign", it is instructive to look at the case when $n=2$. Suppose for all $x,y\in{\bf R}^2$, $$|x_1y_1+x_2y_2|\leq\sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}. $$
Given $a=(a_1,a_2)$ and $b=(b_1,b_2)$, one wants to prove that
$$
|a_1b_1|+|a_2b_2|\leq |a|\cdot|b|. 
$$
If $(a_1b_1)(a_2b_2)\geq 0$, then
$$
|a_1b_1|+|a_2b_2|=|a_1b_1+a_2b_2|\leq |a|\cdot|b|.
$$
If
$(a_1b_1)(a_2b_2)<0$, then
$$
|a_1b_1|+|a_2b_2|=|a_1b_1-a_2b_2|\leq |a|\cdot|b|.
$$
where we use (1) and the fact that $|(b_1,b_2)|=|(b_1,-b_2)|$.
A: Yes, they are not equal but mod of sums is less than or equal to sum of mods and hence it follows. So, second implies the first one for sure.
A: By C-S
$$\sqrt{(x_1^2+x_2^2+...+x_n^2)(y_1^2+y_2^2+...+y_n^2)}=$$
$$=\sqrt{(|x_1|^2+|x_2|^2+...+|x_n|^2)(|y_1|^2+|y_2|^2+...+|y_n|^2)}\geq$$
$$\geq|x_1|\cdot|y_1|+|x_2|\cdot|y_2|+...+|x_n|\cdot|y_n|=$$
$$=|x_1y_1|+|x_2y_2|+...+|x_ny_n|.$$
The equality occurs for $$\left(|x_1|,|x_2|,..., |x_n|\right)||\left(|y_1|,|y_2|,..., |y_n|\right)$$.
In another hand, by C-S
$$\sqrt{(x_1^2+x_2^2+...+x_n^2)(y_1^2+y_2^2+...+y_n^2)}\geq|x_1y_1+x_2y_2+...+x_ny_n|.$$
the equality occurs for $(x_1,x_2,...,x_n)||(y_1,y_2,...,y_n)$.
