I need to prove the following inequality:
$|\sqrt{|a|+1}-\sqrt{|b|+1}|\le \dfrac{|a-b|}{2}$
I started from the LHS and multiplied it by $\sqrt{|a|+1}+\sqrt{|b|+1}$ and obtained the following result:
$\left|\dfrac{\left(\sqrt{|a|+1}-\sqrt{|b|+1}\right)\cdot \left(\sqrt{|a|+1}+\sqrt{|b|+1}\right)}{\left(\sqrt{|a|+1}+\sqrt{|b|+1}\right)}\right|$
Then I used the identity $a^2-b^2 = (a+b)(a-b)$, splitted the absolute value to the nominator and denominator and obtained
$\dfrac{\left||a|-|b|\right|}{\left|\sqrt{|a|+1}+\sqrt{|b|+1}\right|}$
Now using the triangle inequality and the last step I could did without needing help I got to the following result:
$\dfrac{\left||a|-|b|\right|}{\left|\sqrt{|a|+1}+\sqrt{|b|+1}\right|} \le \dfrac{|a-b|}{\left|\sqrt{|a|+1}+\sqrt{|b|+1}\right|}$
I thought that reducing the denominator can help me but couldn't know how to do so..
Lemma 1: If $x\geq y\geq 1,$ then $\left\lvert\sqrt{x} - \sqrt{y}\right\rvert \leq \frac{\left\lvert x-y\right\rvert}{2}.\ $ Proof: $\sqrt{x} + \sqrt{y}\geq 2.$ Thus, $\left\lvert\sqrt{x} - \sqrt{y}\right\rvert= \frac{\left\lvert x-y\right\rvert}{\sqrt{x} + \sqrt{y}} \leq \frac{\left\lvert x-y\right\rvert}{2}.\blacksquare$
Lemma 2: $\vert \vert a \vert - \vert b \vert \vert \leq \vert a - b \vert.\ $ Proof: Put $ u = a-b $ into the triangle inequality $\vert u + b \vert \leq \vert u \vert + \vert b \vert,$ and rearrange to get $ \vert a \vert - \vert b\vert \leq \vert a - b \vert = \vert b - a \vert.\ $ If $\ \vert a \vert \geq \vert b \vert,\ $ then $ \vert \vert a \vert - \vert b \vert \vert = \vert a \vert - \vert b\vert \leq \vert a - b \vert.\ $ Else if $\ \vert a \vert < \vert b \vert,\ $ then $\ \vert \vert a \vert - \vert b \vert \vert = \vert \vert b \vert - \vert a \vert \vert = \vert b\vert - \vert a\vert \leq \vert b - a \vert = \vert a - b \vert.\blacksquare$
Therefore,
$$ \left\lvert\sqrt{\vert a \vert + 1} - \sqrt{ \vert b \vert+ 1}\right\rvert \leq \frac{\vert \vert a \vert - \vert b\vert \vert}{2} \leq \frac{\vert a-b \vert}{2},$$
where the first inequality arose from putting $x=\vert a \vert +1,\ y = \vert b \vert +1\ $ into Lemma 1, and the second inequality is Lemma 2.