Why is $|x| \leq \sqrt{x^2 + y^2}$ for all real $x$ and $y$?

Question

Consider $(x, y) \in \mathbb{R}^2$. From the triangle inequality, we have $$||(x, y)|| \leq ||(x, 0)|| + ||(y, 0)||$$ which means that $$\sqrt{x^2+y^2} \leq \sqrt{x^2} + \sqrt{y^2} = x+y.$$

Whether or not that's relevant, I'd like to conclude that $$|x| \leq \sqrt{x^2 + y^2}$$ for all real $x$ and $y$, and Wolfram Alpha says that I can; but, I don't see how to do this. Could someone please show why this last inequality is true for all real $x$ and $y$?

$\sqrt{x^2+y^2}\le x+y$ is false when $x$ and $y$ are negative. — Angina Seng, Jun 11 '19 at 06:11
I don’t know why so many people don’t know what the symbol $\sqrt{,}$ means. The idea that $\sqrt{x^2}=\pm x$ is complete nonsense: the symbol $\sqrt{x^2}$ refers to a number and of course a number can’t equal both $x$ and $-x$ (when $x$ is nonzero). The symbol means “take the nonnegative square root.” It never makes any sense to write $a=\pm 2$, no matter what you take $a$ to be. Don’t ever write “something$=\pm$ something” again! — symplectomorphic, Jun 11 '19 at 06:17

score 2 · Accepted Answer · answered Jun 11 '19 at 06:32

2

$x,y \in \mathbb{R}$

1)$|x|=\sqrt{x^2};$

2) $x^2 \le x^2+y^2$, since $y^2 \ge 0.$

3) Square root is an increasing function.

Hence

4)$ |x| \le \sqrt{x^2+y^2}.$

answered Jun 11 '19 at 06:32

Peter Szilas

20,344
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score 1 · Answer 2 · answered Jun 11 '19 at 06:12

1

Use $y^2\ge0$ to conclude that $x^2\le x^2+y^2$ and then take square roots.

answered Jun 11 '19 at 06:12

Angina Seng

158,341

Why is $|x| \leq \sqrt{x^2 + y^2}$ for all real $x$ and $y$?

2 Answers2