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

Question

Any hints on how I can prove the inequality:

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

Thank you.

Answer 1

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.

Answer 2

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 $b.$

Answer 3

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.

Answer 4

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.

Answer 5

Suppose $x\neq 0\wedge 0<\frac{y}{x}<1$. Then $\frac{y}{x}\leq\sqrt{\frac{y}{x}}\Rightarrow 1-2\sqrt{\frac{y}{x}}+\frac{y}{x}\leq 1-\frac{y}{x}\Rightarrow 1-\sqrt\frac{y}{x}\leq \sqrt{1-\frac{y}{x}}\Rightarrow$ $\sqrt x-\sqrt y\leq \sqrt{x-y}$

Answer 6

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 $$f'(x)=\frac{1}{2\sqrt{x-1}} - \frac{1}{2\sqrt{x}}\ge 0$$

Answer 7

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.

Answer 8

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.