Prove that for any $x\geq0$ and $y \ge 0$ we have: $|\sqrt{x} − \sqrt{y}| \le \sqrt{|x − y|}.$

Question

Prove that for any $x\geq0$ and $y \ge 0$ we have: $$|\sqrt{x} − \sqrt{y}| \le \sqrt{|x − y|}.$$

$$|\sqrt(x) - \sqrt(y)|^2 \le \sqrt{|x-y|}^2$$ $$(\sqrt{x}-\sqrt{y})^2 \le |x-y|$$ $$x - 2\sqrt{xy} + y \le |x-y|$$

How do I take the absolute value separately to prove it?

Answer 1

After squaring of the both sides we need to prove that $$|\sqrt{x}-\sqrt{y}|^2\leq(\sqrt{x}+\sqrt{y})|\sqrt{x}-\sqrt{y}|$$ or $$|\sqrt{x}-\sqrt{y}|\leq\sqrt{x}+\sqrt{y},$$ which is obvious: $$|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}+\sqrt{y}|=\sqrt{x}+\sqrt{y}.$$ Done!

Answer 2

$$x=|x+y-y|\leq y+|x-y| \Rightarrow \sqrt{x} \leq \sqrt{y+|x-y|} \leq \sqrt{y}+\sqrt{|y-x|}$$

$$y=|y+x-x|\leq x+|y-x| \Rightarrow \sqrt{y} \leq \sqrt{x}+\sqrt{|y-x|}$$

Thus $|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|}$

Answer 3

The last inequality you have is symmetric in $x$ and $y$, that is, replacing $x$ with $y$ and vice-versa does not change the inequality. So, without loss of generality, assume $x \geq y$. You can now remove the modulus on the right, and you can continue to solve.

Answer 4

Consider $2$ cases: $$\begin{align} & x<y \Rightarrow \sqrt{x}-\sqrt{y}<0< \sqrt{|x-y|}, \\ & x\ge y \Rightarrow \sqrt{x}-\sqrt{y}\le \sqrt{x-y} \Rightarrow x-2\sqrt{xy}+y\le x-y \Rightarrow y\le x. \end{align}$$