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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?

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    $\begingroup$ 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 $y <0$. $\endgroup$ Jun 20 at 14:46
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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.

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  • $\begingroup$ Right, I forgot to square the terms inside the square root. $\endgroup$
    – soupless
    Jun 20 at 17:37
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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 $x,y\in\Bbb{R}$.

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