Proving the triangle inequality for the $l_2$ norm $\|x\|_2 = \sqrt{x_1^2+x_2^2+\cdots+x_n^2}$ I want to prove the triangle inequality for the l2-norm $\|x\|_2$:
$$\|x\|_2 = \sqrt{x_1^2+x_2^2+\cdots+x_n^2}$$
$$\begin{align}
  \sqrt {\sum\limits_{i = 1}^n (x_i + y_i)^2 }  &\leqslant \sqrt {\sum\limits_{i = 1}^n x_i^2 }  + \sqrt {\sum\limits_{i = 1}^n y_i^2} \cr 
  \sum\limits_{i = 1}^n {(x_i + y_i)}^2  &\leqslant \sum\limits_{i = 1}^n x_i^2  + 2\sqrt {\sum\limits_{i = 1}^n x_i^2} \sqrt {\sum\limits_{i = 1}^n y_i^2 }  + \sum\limits_{i = 1}^n y_i^2 \cr 
  \sum\limits_{i = 1}^n (x_i^2 + 2x_i y_i + y_i^2)  &\leqslant \sum\limits_{i = 1}^n x_i^2 + 2\sqrt {\sum\limits_{i = 1}^n x_i^2 } \sqrt {\sum\limits_{i = 1}^n y_i^2 }  + \sum\limits_{i = 1}^n y_i^2 \cr 
  \sum\limits_{i = 1}^n x_i^2 + \sum\limits_{i = 1}^n y_i^2 + 2\sum\limits_{i = 1}^n x_i y_i  &\leqslant \sum\limits_{i = 1}^n x_i^2 + \sum\limits_{i = 1}^n y_i^2 + 2\sqrt {\sum\limits_{i = 1}^n x_i^2 } \sqrt {\sum\limits_{i = 1}^n {y_i^2} }  \cr 
  \sum\limits_{i = 1}^n x_i y_i  &\leqslant \sqrt {\sum\limits_{i = 1}^n x_i^2 } \sqrt {\sum\limits_{i = 1}^n y_i^2 }  \end{align}$$
After this I squared both sides and multiplied out. It gets really messy, and then I got stuck. I think you have to realize some factorization and then conclude $0\leq \text{some terms}$
1.How do I finish this proof?  All help is greatly appreciated! 

You want to prove that $$\lVert x+y\rVert \leq \lVert x\rVert+\lVert y\rVert$$
We use $\langle x,y\rangle$ to denote the inner (dot) product $$x\cdot y=\sum_{k=1}^n x_ky_k$$
Note that $$\lVert x+y\rVert ^2=\lVert x\rVert^2+2\langle x,y\rangle+\lVert y\rVert ^2$$
Continuing, use the Cauchy-Schwarz inequality, you get  $$\lVert x\rVert^2+2\langle x,y\rangle+\lVert y\rVert ^2\leq \lVert x\rVert^2+2 \lVert x\rVert \cdot \lVert y\rVert +\lVert y\rVert ^2=\left(\lVert x\rVert+\lVert y\rVert\right)^2 $$ and since all is positive, we can obtain your inequality by taking square roots.
The last equation you got is almost $$|\langle x,y\rangle|\leq \lVert x\rVert \cdot \lVert y\rVert $$
which is known as the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality is true for any inner product space, and here you have the canonical proof. Your inequality is a particular case of the Minkowski inequality for $L_p$ norms.
ADD To clear out the proof in Wikiepdia
Recall that the nature of roots of the polynomial $aX^2+bX+c$ are intimately related to its discriminant: $\Delta=b^2-4ac$ which appears in the formula $$X=\frac{-b\pm\sqrt{\Delta}}{2a}$$
If a quadratic polynomial is proven to satisfy $p(X)\geq 0$ then $\Delta \leq 0$: either there is one real root $\Delta=0$ or there are no real roots ($\Delta<0$) because the square root of the negative discriminant will give rise to complex numbers.
