How to resolve this absolute value inequality $|1+x^2|>|x|$? I am stuck trying to find all x that satisfies
$$|1+x^2|>|x|$$
($x=0$ is obvious.)
To provide more context, I had applied the root test on the series 
$$\sum_{n=1}^{\infty} \frac{nx^n}{n^2 + x^{2n}} $$
and set the limit < 1, to force convergence.  
The n's conveniently converged, leaving me with only the inequality left above, involving only x.
Any ideas are welcome.
Thanks,
 A: One may write
$$
|1+x^2|>|x|
$$ 
$$
|x|^2+1>|x|
$$ 
$$
|x|^2-|x|+1>0
$$$$
\left(|x|-\frac12\right)^2+\frac34>0
$$ thus all real values $x$ satisfy the initial inequation.


Concerning the convergence of $\displaystyle \sum_{n=1}^{\infty} \frac{nx^n}{n^2 + x^{2n}}$ one may consider the following cases.
  
  
*
  
*Assume $|x|<1$. 
Then, we have $$\left|\sum_{n=1}^{\infty} \frac{nx^n}{n^2 + x^{2n}}\right|\leq\sum_{n=1}^{\infty}\left| \frac{nx^n}{n^2 + x^{2n}}\right|\leq \sum_{n=1}^{\infty} n|x|^n=\frac{|x|}{(1-|x|)^2}<\infty,$$ the initial series converges absolutely thus converges.
  
*Assume $|x|>1$.  
Then, we have $$\left|\sum_{n=1}^{\infty} \frac{nx^n}{n^2 + x^{2n}}\right|\leq\sum_{n=1}^{\infty}\left| \frac{nx^n}{n^2 + x^{2n}}\right|\leq \sum_{n=1}^{\infty} \frac{n}{|x|^n}=\frac{|x|}{(1-|x|)^2}<\infty,$$ the initial series converges absolutely thus converges.
  
*Assume $x=1$.  
Then, $$\sum_{n=1}^{\infty} \frac{nx^n}{n^2 + x^{2n}}=\sum_{n=1}^{\infty} \frac{n}{n^2 +1},$$ and as $n \to \infty$, $$\frac{n}{n^2 +1}\sim\frac1{n},$$ the initial series diverges.
  
*Assume $x=-1$.  
Then, $$\sum_{n=1}^{\infty} \frac{nx^n}{n^2 + x^{2n}}=\sum_{n=1}^{\infty} \frac{n}{n^2 +1}(-1)^n,$$ the initial series converges conditionally by the alternating test.

A: Note that for all real $x$, $x^2 \ge 0$, hence $1 + x^2 > 0$, and by the AM-GM inequality, $$|1+x^2| = 1+x^2 \ge 2\sqrt{1 \cdot x^2} = 2 |x| \ge |x|.$$
A: The easiest way to see this is to note if $|x| \le 1$ then $|1 + x^2| \ge 1 \ge |x|$, and if $|x| \ge 1$ then $|1 + x^2| > |x^2| \ge |x|$.
A: You can square both terms of the inequality and obtain $$1+2x^2+x^4>x^2$$
This is equivalent to $x^4+x^2+1>0$ which is true for every $x^2$, thus also for all $x$.
You can also see a more empirical approach by consider two cases. When $x$ has absolute value less than 1 the left hand side term is at least 1 as $x^2$ is positive, therefore the term in the modulus is bigger than 1 which is bigger than $x$.
For $x$ bigger in absolute value than 1, the square of it is positive and exceeds $|x|$, so the inequality holds as well. For example: $1.1^2=1.21>1.1$ :)
A: Since $\forall x :1+x^2$ is positive, |$1+x^2$| = $1+x^2$. 
We only need to solve this inequality:
$1+x^2$ > $x$
$x^2-x+1$ has no roots and is positive for every $x$, this means that every $x$ satisfies the inequality. 
