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?
 A: 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.
A: 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 $x,y\in\Bbb{R}$.
