Prove $|\max\{x, y\} - z| \leq |x - z|$ for $x, y, z \in \mathbb{R}$ such that $y \leq z$? I am trying to check if the following 2 elementary inequality claims  are true.
Claim 1:
Let $x, y, z \in \mathbb{R}$ be such that $y \leq z$.
Then the following inequality holds $$|\max\{x, y\} - z| \leq |x - z|$$
To check this I believe we can proceed as follows:
\begin{align}
    |\max\{x, y\} - z|
    & \leq |\max\{x, z\} - z| \tag{since $y \leq z$} \\
    & \leq |x - z|
\end{align}
Where the last line follows since if $x = z$, then $|\max\{x, z\} - z| = |z - z| = 0$. Otherwise if $x \neq z$, then $|x - z| > 0$, so it is always bigger than $|\max\{x, z\} - z|$.
Claim 2:
Let $x, y, z \in \mathbb{R}$ be such that $y \leq z$.
Then the following inequality holds $$|\max\{x, y\} - z| \leq |x - z|$$
I believe that Claim 2 follows from Claim 1, by applying the increasing function $x \mapsto x^{2}$ on $x \geq 0$ to both sides of our Claim 1 result (assuming it is true). Is this logic to prove the Claim 2 result assuming Claim 1 is true, correct? I believe this is true since both Claim 1 and Claim 2 use the same underlying assumptions on $x, y, z$.
Update: Made it clear that the conditions for Claim 1 and Claim 2 are the same, i.e., they are for $x, y, z \in \mathbb{R}$ such that $y \leq z$.
Update 2: I've edited Claim 1 and Claim 2, since I believe I may have caused confusion, given the thoughtful response from @RWPardo.
I'm not sure the above reasoning is correct, or if the inequality is true. I may have missed something. Could anyone please confirm, and then show it more rigorously than I have? If there is a counterexample, how could this type of inequality be made true?
 A: 
First and easy proof of your question:

If $\max\{x,y\} = x$, we have nothing to do, since the inequality holds.  On the other hand, if $\max\{x,y\} = y$, we have $y \geq x.$ This implies that $$-y\leq -x \tag{*}.$$ On the other hand, since $y \leq z,$ we have $z-y=|y-z| = |\max\{x,y\} - z|.$ Hence, $$z-y=|y-z| = |\max\{x,y\} - z|\overset{(*)}{\leq} z - x \leq |x-z|, $$ which ends the proof.

On your second clain:
Your second claim is partially right!
Let $a \geq b \geq 0$, we also have that there exist a $t\geq 0$ such that $a =  b +t$, hence $a^2 =  b^2 +2 t b +t^2. $ Since $s= 2 t b +t^2 \geq 0$, we have that $$a^2 \geq b^2.$$ Hence the function $f: [0,\infty)\rightarrow \mathbb{R}$ such that $f(x) = x^{2}$ is an increasing function.
Why partially right?
Only using that the function $f$ is increasing is not enough. That's because the inequality $|\max\{x,y\} - z| \leq |z-x|$ does not hold for general $x,y,z \in \mathbb{R},$ but only for $y\leq z.$ The good thing is that proving that $|\max\{x,y\} - z| \leq |z-x|$ for $y\leq z$ is equivalent to proving $(\max\{x,y\} - z)^2 \leq (z-x)^2$ for $y\leq z$. The bad thing is that proving the squared inequality is harder than proving the actually wanted inequality. As a way of saying sorry of my previous and first answer (the second edited answer), I will convince you that proving the squared inequality is harder.

Second and “hard” proof of your question using the squared inequality:

Proving the squared inequality:
Let $x,y,z \in \mathbb{R}$ be real numbers, calling $\max \{x,y\} = w$, we have,
\begin{align*}
(w - z)^2 \leq (x-z)^2 \iff &  w^2 - 2 w z + z^2 \leq x^2-2 x z +z^2 \\
\iff &  w^2 - 2 w z + z^2 \leq x^2-2 x z +z^2 \\
\iff &  w^2 - 2 w z \leq x^2-2 x z  \\
\iff &  w^2 - x^2 \leq 2 w z - 2 x z \\
\iff &  (w-x)(w+x) \leq 2z ( w -x ) \tag{**} \\
\end{align*} The only thing that is left to us is to prove that $(w-x)(w+x) \leq 2z ( w -x )$ for all $z \geq y.$ If $w=x$, nothing we have to prove, since $ 0 = (w-x)(w+x) \leq 2z ( w -x ) = 0.$ On the other hand, if $w>x,$ we have $$w = y \tag{***}$$ and $$y>x \tag{****}$$ since $w=\max\{x,y\}$. Now, note that
\begin{align}
w + x \overset{(***)}{=} & y + x \\
\overset{(****)}{<} & y + y \\ 
\overset{y \leq z }{\leq} & 2 z \tag{*****}.
\end{align} Hence, since $w - x>0$, we have $(w + x) (w - x) \leq 2z (w - x), $ which proves $(**)$ and, hence, your desired inequality. It were harder, right?

Answering questions in the comments:

Before answering here, keep calm. I'm not aware of all your background, but I can clarify some things first. It's not a 7-headed boss in a hard game.
1)  Is everything in my first claim right?
Unfortunately, no. In the passage $$|\max\{x, y\} - z| \leq |\max\{x, z\} - z| \tag{since $y \leq z$},$$ you seem to need to use that the modulus function increases, but it does not. For example, take $x=1$, $y=2$, and $z=3$, we have that $z\geq y$, but it does not hold that $$|\max\{x, y\} - z| \leq |\max\{x, z\} - z|,$$ since the right side is zero and the left side, in modulus, equal to 1.
2) If the first claim was true, my second claim would be true?
Yes, I agree with this, but it is not true, as said in 1.
3) You said:

And is my understanding correct that you have proved a much stronger clai, i.e., $(\max\{x,y\}−z)^2 \leq (x−z)^2, \forall x,y,z\in \mathbb{R}$, with no restriction that $y \leq z$?

No, I have used that $y \leq z$ in the inquality $(*****).$
4) You said:

Also the second inequality does quickly follow from the first. I'm convinced now since $x \mapsto x^2$ is an increasing map for each $x \geq 0$. Since our first inequality is for non-negative terms (assuming x,y,z such that $y\leq z$), we can square both sides and conclude. @fleablood confirmed this reasoning is correct below. I'm still not sure of the reasoning behind your more complicated proof.

I can't say what @fleablood confirmed. I can confirm that the function $x \mapsto x^2$ restricted to $[0,\infty)$ is an increasing function. On the other hand, as I have said before, only using this is not enough for proving your result. You will need to use that $y \leq z$ and first claim is wrong, as I said before.
5) Why the second proof is harder?
This is because we can't assume the $$|\max\{x,y\} - z| \leq |x-z|,$$ for $y\leq z,$ since this is the desired result. Assuming the conclusion to prove itself is called a Circular reasoning. So, if we want to prove the desired inequality, and we can't use the conclusion, we would need to work out the things first. That's what I did.
A: Your first line doesn't work
If $y \le z$ then $\max(x,y) \le z$ and $\max(x,y) - z \le \max(x,z) - z$ but that does not mean that $|\max(x,y) - z| \le |\max(x,z)-z|$
Counter example:  $x=1$ and $y= 2$ and $z=57493085940386403869043$.  Then $\max(x,y) - z = 2 - 57493085940386403869043 = -57493085940386403869041 < 0 = \max(x,z) -z$ but $|\max(x,y)-z| = +57493085940386403869041 > 0 =|\max(x,z)- z|$.
....
I'd consider the two cases $\max{x,y} < z$ and $z \le \max{x,y}$.
If $\max {x,y} < z$ then $x < z$ and so we have $|\max(x,y)-z| = z-\max(x,y)$ and $|x-z| = z-x$.  Ans as $\max(x,y) \ge x$ then $z-\max(x,y) \le z - x$.
And if $z \ge \max{x,y}$ then $|\max(x,y)-z| = \max(x,y) -z$ and $|x-z|$ ... well, we don't know how $x$ compares with $z$ so we can consider cases $x < y$ then $\max(x,y) = y \le z$ so $x < z$ so $|x-z|=x-z$ ans we have $\max(x,y)-z \ge x-z$.
And if $z\ge \max{x,y}$ and $x\ge y$ then $\max(x,y)=x$ and $|\max(x,y)-z| = |x-z|$.
....
As for the second claim:
Yes... if $a< b \implies f(a) < f(b)$ then as we have shown $|\max(x,y)-z| \le |x-z|$ we can conclude $f(|\max(x,y)-z|) \le f(|\max(x,y)-z)$.
But is $x \mapsto x^2$ an increasing function?  Note: $-3 < 2$ but $(-3)^2 = 9 > 4 = 2^2$.  On which conditions is $x\mapsto x^2$ increasing? You need to take care of that.
(Also, maybe you need to confirm that $|M|^2 = M^2$ but... I'll grant you that one.)
