Let $x_1,x_2,\ldots,x_n$ be arbitrary real numbers. Prove the inequality $$\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots + \frac{x_n}{1 + x_1^2 + \cdots + x_n^2} < \sqrt{n}$$
I found a solution here which says:
$$\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2} < \sqrt{n}$$ $$ \iff \left(\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\right) ^2< n$$ but we have: $$ \left(\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\right)^2 \le n\left(\frac{x^2_1}{(1+x_1^2)^2} + \frac{x^2_2}{(1+x_1^2 + x_2^2)^2} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_n^2)^2}\right)$$ $$ \le n\left(\frac{x^2_1}{1\cdot (1+x_1^2)} + \frac{x^2_2}{ (1+x_1^2)(1+x_1^2 + x_2^2)} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_{n-1}^2)(1 + x_1^2 + \cdots + x_n^2)}\right)=n\left(1-\frac{1}{1 + x_1^2 + \cdots + x_n^2}\right)<n$$
My query is that they claimed $$ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2} < \sqrt{n}$$ $$\iff \left(\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\right) ^2< n$$ but the LHS of the first inequality need not to be positive. We know $x^2$ is not strictly increasing in all reals, then how did they conclude the second line from the first one?
