Proving the triangle inequality for the $l_2$ norm $\|x\|_2 = \sqrt{x_1^2+x_2^2+\cdots+x_n^2}$

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.