Proof that $|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$ Any hints on how I can prove the inequality:

$$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$$

Thank you.
 A: Without loss of generality $x\geq y$. Then squaring both sides yields
$$\sqrt{x}-\sqrt{y}\leq\sqrt{x-y}\iff x-2\sqrt{xy}+y\leq x-y.$$
The $x$ cancels; taking $y$ to the left and $2\sqrt{xy}$ to the right yields that the inequality is true if and only if the following inequality is true:
$$2y\leq 2\sqrt{xy}.$$
Now note that $xy\geq yy=y^2$, so
$$y=\sqrt{y^2}\leq\sqrt{xy},$$
as desired.
A: HINT:
For $a,b \ge 0,$
$$0 \le \sqrt{ab}$$
$$a+b \le a+ 2\sqrt{ab}+b=(\sqrt{a}+\sqrt{b})^2$$
$$\sqrt{a+b} \le \sqrt{a}+\sqrt{b}$$
Choose suitable values for $a$ and $b.$
A: Hint: Work backwards.
Try squaring both sides
or
multiplying $\sqrt{x} + \sqrt{y}$ on both sides.
Note: You can square both sides because they are nonnegative.
A: There are three cases to consider. The first case is $0 \le y < x$. For this case, since $\sqrt x $ and $\sqrt y $ are increasing functions, your question reduces to proving that $$\sqrt x  - \sqrt y  \le \sqrt {x - y} $$, which by squaring (noticing the positive sign of both sides) and simplification is proving that $y \le \sqrt {xy} $ in this case, which is obvious ($\sqrt y  \le \sqrt x $ follows from our initial assumption $x > y$). The second case is when $x=y$ which is essentially proving that $0 \le 0$. The third case is also similar to the first case.
A: Suppose $x\neq 0\wedge 0<\frac{y}{x}<1$. Then 
$\frac{y}{x}\leq\sqrt{\frac{y}{x}}\Rightarrow 1-2\sqrt{\frac{y}{x}}+\frac{y}{x}\leq 1-\frac{y}{x}\Rightarrow 1-\sqrt\frac{y}{x}\leq \sqrt{1-\frac{y}{x}}\Rightarrow$
$\sqrt x-\sqrt y\leq \sqrt{x-y}$
A: WLOG $0<y\le x$, and
the inequality is equivalent to
$$\sqrt x-1\le\sqrt{x-1}$$
for $x\ge 1$.But
$$f(x)=\sqrt{x-1}-(\sqrt x-1)$$ is increasing in $[1,+\infty)$ because
$$f'(x)=\frac{1}{2\sqrt{x-1}} - \frac{1}{2\sqrt{x}}\ge 0$$
A: Case 1:  If $x = y$, both sides equal zero.  
Case 2:  If $x \neq y$, both sides of the inequality are positive, so we can square both sides of 
$$|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|}$$
to obtain
$$|\sqrt{x} - \sqrt{y}|^2 \leq |x - y| = |(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})| = |\sqrt{x} + \sqrt{y}||\sqrt{x} - \sqrt{y}|$$
Divide both sides of the inequality by $|\sqrt{x} - \sqrt{y}|$ to obtain
$$|\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} + \sqrt{y}|$$
Since both sides of the inequality are positive, we can square both sides to obtain
$$x - 2\sqrt{xy} + y \leq x + 2\sqrt{xy} + y$$
Cancelling $x + y$ yields
$$-2\sqrt{xy} \leq 2\sqrt{xy}$$
Dividing both sides of the inequality by $-2\sqrt{xy}$ yields
$$0 \geq -1$$
Since the steps are reversible, the original inequality holds.
A: Squaring twice, we can use $a^2-b^2=(a-b)(a+b)$ to arrive at:
$$
(\sqrt{x}-\sqrt{y})^4\leq(x-y)^2\iff0\leq(x-y)^2-(x-2\sqrt{x}\sqrt{y}+y)^2=4\sqrt{x}\sqrt{y}(\sqrt{x}-\sqrt{y})^2
$$
which is clearly true. There's no need to do cases.
