# Why does the triangle inequality seem to be false when rewritten based on $|x| = \sqrt{x^{2}}$?

The triangle inequality is $$|x + y| \leq |x| + |y|.$$

Also, we know that $$|x| = \sqrt{x^{2}}$$. Then,

\begin{align*} \sqrt{x + y} &\leq \sqrt{x} + \sqrt{y} \\ x + y &\leq x + y + 2\sqrt{xy} \\ 0 &\leq 2\sqrt{xy} \\\\ \sqrt{xy} &\geq 0 \end{align*}

Now, we can see that the inequality $$|x + y| \le |x| + |y|$$ holds for real $$x$$ and $$y$$, but $$\sqrt{xy} \geq 0$$ does not for $$x < 0, y > 0$$ or $$x > 0, y < 0$$.

What seems to be the problem? Is it the statement $$|x| = \sqrt{x^{2}}$$ or is it much more than that?

• If $x < 0$ and $y > 0$ then $\sqrt{xy}$ is not defined for real numbers since $xy$ will be negative. The same is true for $x > 0$ and $y <0$. Jun 20 at 14:46

The statement you have written using $$|x|=\sqrt {x^2}$$ is wrong. The correct statement would be: $$\sqrt {(x+y)^2}\leq \sqrt {x^2}+\sqrt {y^2}$$ which, indeed, does hold for all real $$x,y$$, as can be verified by squaring and simplifying.
The problem is not with writing $$|x|=\sqrt{x^2}$$: that statement is true as long as $$x$$ is a real number. But the correct translation of $$|x+y|\le|x|+|y|$$ is $$\sqrt{(x+y)^2} \le \sqrt{x^2} + \sqrt{y^2} \tag{*}\label{*} \, ,$$ not what you wrote. To prove that $$\eqref{*}$$ is true, we can use the fact that if $$a$$ and $$b$$ are nonnegative, then $$a\leq b \iff a^2\le b^2$$. Hence, \begin{align} & \sqrt{(x+y)^2} \le \sqrt{x^2} + \sqrt{y^2} \\[4pt] \iff & (x+y)^2 \le x^2+y^2+ 2\sqrt{x^2}\sqrt{y^2} \\[4pt] \iff & x^2+y^2+2xy \le x^2+y^2 + 2|x||y| \\[4pt] \iff & 2xy \le 2|x||y| \\[4pt] \iff & xy \le |x||y| \\[4pt] \iff & xy \le |xy| \, . \end{align} Since the final line is true for all $$x,y\in\Bbb{R}$$, the first line must also be true for all $$x,y\in\Bbb{R}$$.