How can I prove this inequality about square roots and absolute value? How can I prove that $$\left|x^{1/2}-y^{1/2}\right|\leq\left|x-y\right|^{1/2}$$ for all $x,y \geq 0$
 A: First, assume $x\geq y$. Then we can lose the absolute value signs. We now have
$$
\sqrt{x} - \sqrt{y} \leq \sqrt{x-y}\\\\
(\sqrt{x} - \sqrt{y})^2 \leq x-y\\\\
x -2\sqrt{xy} + y \leq x - y \\\\
0 \leq 2\sqrt{xy} - 2y
$$
Since we assumed $x \geq y \geq 0$, the last inequality holds, and the steps are completely reversible with this asumption, so the first inequality has to be true.
A: $x=a^2,y=b^2$ and square both sides. Look more familiar yet? Further hint: Look for a common factor!
A: Without loss of generality assume $x \geq y$. If $y=0$, the inequality is obvious. Hence, lets take $\vert y \vert \neq 0$. Divide by $\vert y \vert$ and let $t=\dfrac{x}y$. We then need to prove that for $t\geq 1$(because $x\geq y)$
$$\sqrt{t} -1 \leq \sqrt{t-1} \tag{$\star$}$$
Consider the function $f(t) = \sqrt{t-1} - \sqrt{t}$ for $t>1$. We have (for $t\geq 1)$
$$f(t) = - \dfrac{\sqrt{t}-\sqrt{t-1}}{\sqrt{t}+\sqrt{t-1}} (\sqrt{t} + \sqrt{t+1}) = - \dfrac1{\sqrt{t} + \sqrt{t-1}} \geq - 1$$
This means $$\sqrt{t-1} - \sqrt{t} \geq -1 \implies \sqrt{t} - 1 \leq \sqrt{t-1} \implies (\star)$$
A: Just remember that $\lvert u \rvert^2 = u^2$, and also that $(u^{1/2})^2 = \lvert u \rvert$
