If you study the convergence on the interval $[1,\infty)$, then you're interested in $\sup\limits_{x\in[1,\infty)} |f_n(x)-f(x)|$. Since $\frac1n\notin[1,\infty)$, this is not what you computed in the first part.
I guess that if you have a look at your computations again, you'll find out that r'_n(x)<0 for $x>\frac1n$, which implies that the function $r_n(x)=f_n(x)-f(x)$ is decreasing on $[1,\infty)$. (In fact, it is decreasing already on a bigger interval $[1/n,\infty)$.)
Thus the maximal possible value is at the leftmost point of this interval, which is $r_n(1)=f_n(1)=\frac{2n}{1+n^2}$. This converges to zero.
So your first approach is fine, the only thing is you have to find the supremum of $|f_n(x)-f(x)|$ on the same interval, for which you want to test the uniform convergence.
EDIT: The rest is about the first version of the post, without the condition $x\in[1,\infty)$. I think the best thing is to keep it here too.
In your second solution you used the inequality which is not valid. You argued that: $\frac1{nx}+nx \ge n \Rightarrow \frac2{\frac1{nx}+nx} \le \frac 2n.$
In fact you only have $\frac1{nx}+nx \ge 2 \Rightarrow \frac2{\frac1{nx}+nx} \le \frac 22=1,$ which is consistent with your first solution.
There are many ways to see that for $t>0$ we have $t+\frac1t\ge 2.$ (This is the inequality I used above for $t=nx$.)
E.g. you can study the derivative of the function $f(t)=t+\frac1t$. Quite simple way to see this is using AM-GM inequality, which gives you: $\frac{t+\frac1t}2\ge \sqrt{t\cdot\frac1t}=1.$
It is known that we get the equality in AM-GM only when the summands are equal, which is in our case $t=\frac1t$, i.e. $nx=t=1$ and $x=\frac1n$; just the same as you found in your first approach.