Need better proof: $\mid \sqrt{a} - \sqrt{b}\mid \le \sqrt{\mid a - b \mid} $ , if $a \ge 0 \thinspace and \thinspace b\ge0$.

Question

Prove the inequality: $\mid \sqrt{a} - \sqrt{b}\mid \le \sqrt{\mid a - b \mid} $ , if $a \ge 0 \thinspace and \thinspace b\ge0$.

This is a homework and I have solved it with 2 method:

1st method: Prove with the graphic of 2 squres overlap, whose ares are a and b;

2nd method: Separate $a \ge 0 , b \ge 0$ into 5 conditions below and prove $\mid \sqrt{a} - \sqrt{b}\mid \le \sqrt{\mid a - b \mid} \thinspace $ holds for each condition. Druing the proof, I used triangle inequality and average inequality. $ (1)\thinspace\thinspace a=0 ,b=0 \thinspace\thinspace(2)\thinspace \thinspace a=0 ,b>0 \thinspace\thinspace(3)\thinspace \thinspace a>0, b=0 \thinspace\thinspace(4)\thinspace \thinspace a>0 ,b>0 ,a\neq b \thinspace\thinspace(5)\thinspace \thinspace a>0, b>0 , a=b \thinspace \thinspace$

I think my solution is neither simple nor efficient.

Can you give me a better proof ? Think you！

Answer 1

Note that

$$|\sqrt a-\sqrt b|=\frac{|a-b|}{\sqrt a+\sqrt b}=\color{blue}{\frac{\sqrt{|a-b|}}{\sqrt a+\sqrt b}}\sqrt{|a-b|}\tag 1$$

$$|a-b|\leq a+b\leq a+b+2\sqrt{a}{\sqrt b}=(\sqrt a+\sqrt b)^2\tag 2$$

The first inequality in $(2)$ is due to triangular inequality.

From $(2)$, it follows that$\sqrt {|a-b|}\leq \sqrt a+\sqrt b$.

So the blue colored part in $(1)$ is $\leq 1$.

Answer 2

Square both hands to get $$a + b - 2\sqrt{ab} \leq |a - b|.$$ If $a \geq b$, then the right hand side becomes $a - b$, and hence the inequality can be rewritten as $$(a + b) - (a - b) \leq 2\sqrt{ab}.$$ This can be simplified to $$2b \leq 2\sqrt{ab},$$ which should be obviously true due to our assumption that $a \geq b$.

The case where $a < b$ can be solved similarly.

Answer 3

I usually think of this as a consequence of the concavity of the square root by writing it as $$ \sqrt a -\sqrt b \leq \sqrt{a-b} - \sqrt 0 $$ when $a\geq b$. Indeed, the derivative of a concave function $u$ is decreasing and so $$ \int_b^{b+(a-b)} u' \leq \int_0^{a-b} u' $$ This is easy to see on a drawing: "jumping of a size $a-b$ makes a bigger effect close to $0$".

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Notice also the advantage that this work for any concave function, so for example $x^\theta$ with $\theta \in[0,1]$.

Answer 4

We take $a \geq 0$ and $b \geq 0$.

Then $\sqrt{a}$ and $\sqrt{b}$ are well-defined.

We wish to prove: $$ | \sqrt{a}- \sqrt{b} | \leq \sqrt{ | a - b |} \tag{1} $$

Clearly, proving (1) is equivalent to proving the following:

$$ | \sqrt{a}- \sqrt{b} |^2 \leq [\sqrt{ | a - b |}]^2 $$ or $$ | \sqrt{a} - \sqrt{b} |^2 \leq | a - b | $$ or $$ a + b - 2 \sqrt{a b} \leq | a - b | \tag{2} $$

We have two cases to consider:

Case 1: When $a \geq b$

In this case, $| a - b | = a - b$.

Then (2) can be simplified as $$ a + b - 2 \sqrt{a b} \leq a - b $$ or $$ 2 b \leq 2 \sqrt{a b} $$ or $$ b \leq \sqrt{a b} \tag{3} $$

Since by assumption in Case 1, $a \geq b$, it is immediate that $$ \sqrt{a b} \geq \sqrt{b \times b} = \sqrt{b^2} = b. $$

Thus, we have established that (3) is true.

Case 2: When $a < b$.

In this case, $| a - b | = b - a$.

Then (2) can be simplified as $$ a + b - 2 \sqrt{a b} \leq b - a $$ or $$ 2 a \leq 2 \sqrt{a b} $$ or $$ a \leq \sqrt{a b} \tag{4} $$

Since by assumption in Case 2, $b > a$, it is immediate that $$ \sqrt{a b} > \sqrt{a \times a} = \sqrt{a^2} = a. $$

Thus, we have established that (4) is true.

The proof is complete by combining Cases 1 and 2. $\ \ \ $ $\blacksquare$