1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $(\|x\| + \|y\|)^2 - \|x + y\|^2 \geq 0$".

$\endgroup$
  • $\begingroup$ Thank you so much! I sometimes feel so frustrated getting stuck with wrong approach thinking this must be the right approach. I really hope getting intuition for solving proof problems will eventually come with practice. How did you come up with the idea so fast if I may ask? I really would appreciate some advices on getting better at proving these proof problems. Thank you so much again. $\endgroup$ – user3000482 Mar 29 '17 at 5:48
  • 1
    $\begingroup$ @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. $\endgroup$ – Omnomnomnom Mar 29 '17 at 5:57
  • 1
    $\begingroup$ 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 $\|x\|^2 = \langle x,x \rangle$. $\endgroup$ – Omnomnomnom Mar 29 '17 at 5:58
  • 1
    $\begingroup$ 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 $n = 2$. $\endgroup$ – Omnomnomnom Mar 29 '17 at 6:02
  • $\begingroup$ Thank you so much for additional comments. I really appreciate all the comments and I can't thank you enough. $\endgroup$ – user3000482 Mar 29 '17 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.