0

Basically, the inequality says that the norm of a vector is always less than or equal to the sum of the absolute value of its components.

And I know that the norm is defined as:

$$\|(a,b)\|=\sqrt{a^2+b^2}$$

and the absolute value as:

$$|a|=\sqrt{a^2}$$

But I don't know how to relate both equations and prove the inequality.

Xander Henderson
  • 29,772
Javier Poncet
  • 5
  • 2
    For a start: square the inequality. – Michael Hoppe Oct 14 '18 at 22:29
  • 1
    Do you know the triangle inequality? If so, write $(a,b) = (a,0)+(0,b)$ and apply it. – Clement C. Oct 14 '18 at 22:36
  • @XanderHenderson No, that's not the same thing. Read the question. – Clement C. Oct 14 '18 at 22:40
  • @ClementC. I have read the question. As far as I can tell, the asker is asking for a proof of the triangle inequality. If you think that the asker is wanting something else, perhaps you or the asker could elucidate? It is also worth noting that answers to the the linked question adequately answer this question, e.g. hunminpark's answer is essentially identical to cansomeonehelpmeout's answer here. – Xander Henderson Oct 14 '18 at 22:44
  • @XanderHenderson No, the proof is similar to that of the triangle inequality, but this is not proving the same thing. The triangle inequality is about the norm of the sum of two vectors vs. the sum of the norms of the two vectors. This is about the (Euclidean) norm of a vector vs. the sum of absolute values of its components. – Clement C. Oct 14 '18 at 22:47
  • @ClementC. So, it's the triangle inequality as applied to the vectors $(a,0)$ and $(0,b)$. – Angina Seng Oct 15 '18 at 01:41
  • @LordSharktheUnknown Erm, yes. Which is what my first comment said, if you have a look at it. It's a direct application of the triangle inequality, but it is not the triangle inequality itself. Xander Handerson's comment states that "the asker is asking for a proof of the triangle inequality." This is not the case (although a proof that the Euclidean norm satisfies the triangle inequality is very similar to a direct proof of the OP's inequality). – Clement C. Oct 15 '18 at 01:42

2 Answers2

2

Simply use the triangle inequality: $\;(a,b)=(a,0)+(0,b)$, so $$\|(a,b)\|\le\|(a,0)\|+\|(0,b)\|=|a|+|b|.$$

Bernard
  • 175,478
1

Squaring gives: $$ ||(a,b)||\leq|a|+|b|\iff (||(a,b)||)^2\leq(|a|+|b|)^2\iff 0\leq 2|a||b|. $$

yurnero
  • 10,505