proving that $f_n$ doesn't uniformly converge I am trying to prove that $$fn(x)=(nx)/(1+n^2x^4)$$ doesn't uniformly converge in [0,1]
For that I'm looking for K a real positive fixed number, m,n $\in \mathbb{N}$ which achieve the following inequality:
$$|nx/(1+n^2x^4) - mx/(1+m^2x^4)| \ge K. $$ 
Why I'm doing this?
well because we learnt that fn(x) uniformly converge when for every epsilon > 0 there is N a natural number S.t for every m,n > N and every x in my field the following is true:
|fn(x)-fm(x)| <= epsilon
So I wasn't successful solving this and uploaded the following question:
https://math.stackexchange.com/posts/3685851

Given $n \in \mathbb{N}$, I'm looking for $m \in \mathbb{N}$ which
  achives the following:
Note: $m$ is expected to be related to $n$. For example $m=2n$....
$$|n/(1+n^2) - m/(1+m^2)| \ge K. $$ $K > 0$, $K$: constant. An
  Example: $K = 0.5$

and it seems that people are struggling with it, am I doing something wrong here?

With $x:=\frac1{\sqrt n}$ check that
$$f_n(x)=\frac{nx}{1+n^2x^4}=\frac{\sqrt n}2$$
and
$$f_{2n}(x)=\frac{2nx}{1+4n^2x^4}=\frac{2\sqrt n}5$$
so that
$$
\left|f_n(x)-f_{2n}(x)\right| =\frac{\sqrt n}{10}
$$
So given $N\ge1$ you can take $n=N$, $m=2N$, $x=\frac1{\sqrt N}$ to satisfy your Cauchy criterion. For the value of $K$ you can take $K=\frac{1}{10}$.
The question you uploaded previously received no answer because you didn't translate your Cauchy criterion correctly: you forced $x=1$, which doesn't  yield any $m, n$ that satisfy.
