# Need help on proving triangle inequality for norm.

I need to prove $\lVert \mathbf{x} + \mathbf{y}\rVert \leq \lVert \mathbf{x} \rVert+\lVert\mathbf{y}\rVert$, where $\mathbf{x},\mathbf{y}\in\mathbb{R}^n$. The norm is given as $\lVert\mathbf{x}\rVert = \sqrt{\langle\mathbf{x},\mathbf{x}\rangle}$ where $\langle\mathbf{x},\mathbf{y}\rangle = x_1y_1+x_2y_2+\cdots+x_ny_n$, the scalar product of two vectors.

The book says I need to use the Cauchy-Schwarz Inequality, which is $|\langle\mathbf{x},\mathbf{y}\rangle| \leq \lVert\mathbf{x}\rVert\cdot\lVert\mathbf{y}\rVert$.

This is how far I got:

$\lVert\mathbf{x}+\mathbf{y}\rVert = \sqrt{x_1^2+\cdots+x_n^2+y_1^2+\cdots+y_n^2+2(x_1y_1+\cdots+x_ny_n)}$

$\leq \sqrt{x_1^2+\cdots+x_n^2} + \sqrt{y_1^2+\cdots+y_n^2}+\sqrt{2(x_1y_1+\cdots+x_ny_n)}$

$= \lVert\mathbf{x}\rVert + \lVert\mathbf{y}\rVert + \sqrt{2(x_1y_1+\cdots+x_ny_n)}$

I think I need to somehow get rid of $\sqrt{2(x_1y_1+\cdots+x_ny_n)}$ from the last inequality but I can't figure out how. $(x_1y_1+\cdots+x_ny_n) < 0$, then I think I can neglect $\sqrt{2(x_1y_1+\cdots+x_ny_n)}$. But would this be also true even if $(x_1y_1+\cdots+x_ny_n) \geq 0$?

I can't think of a way to use Cauchy-Schwarz inequality to finish my proof. Am I on right track on proving this or should I have taken different approach?

I would take a different approach here. The usual approach is to start with $$\|x+y\|^2 = \langle x+y, x+y \rangle = \|x\|^2 + \|y\|^2 + 2\langle x,y \rangle$$ Compare this to $$(\|x\| + \|y\|)^2 = \|x\|^2 + \|y\|^2 + 2 \|x\| \, \|y\|$$
To package this all nicely: "it suffices to prove that $(\|x\| + \|y\|)^2 - \|x + y\|^2 \geq 0$".
• @user3000482 It definitely comes with practice. I've seen this question a lot of times, so I just know the answer. That being said, there are certain intuitions here that are helpful: first, using the inequality $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$ "loses precision" to the inequality. If you're trying to prove a specific inequality (as opposed to constructing an arbitrary upper bound), you should aim for the smallest amount of intermediate inequalities that you can get away with. – Omnomnomnom Mar 29 '17 at 5:57
• Another handy piece of intuition: the dot product (and any inner product) is an excellent tool to have. If $\|x\|$ comes from an inner product, try to use $\|x\|^2$ in proofs whenever possible, since $\|x\|^2 = \langle x,x \rangle$. – Omnomnomnom Mar 29 '17 at 5:58
• Another piece of intuition: how would you prove this for the complex numbers? $$|z + w| \leq |z| + |w|$$ somehow, the process I demonstrate seems more intuitive; squaring both sides seems natural. However: with this proof, we're effectively proving our statement for $n = 2$. – Omnomnomnom Mar 29 '17 at 6:02