$\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $\left|x\right|$ Im trying to prove that $f_{n}\left(x\right)=\sqrt{x^{2}+\frac{1}{n}}$  converges uniformly to $f(x) = \left|x\right|$ in $[-1,1]$. 
So for evary $\varepsilon$ exists $N \in \mathbb{N}$ s.t for all $n>N$ and for all $x \in [-1,1]$   $\left|f_{n}\left(x\right)-f\left(x\right)\right| \le \varepsilon$
And this how i tried to continue : 
$\left|f_{n}\left(x\right)-f\left(x\right)\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\left|x\right|\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|=\left|\frac{\left(\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right)\left(\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}\right)}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{x^{2}+\frac{1}{n}-x^{2}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{\frac{1}{n}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|$
Which lead me to the same problem of $\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|$
any advice ?

If $a,b\ge 0,$ then $\sqrt {a+b} \le \sqrt a + \sqrt b.$ Proof: Square both sides. We conclude
$$\sqrt {x^2+1/n} \le \sqrt {x^2}+\sqrt {1/n},$$
which will lead to your result. 
