Short proof that $\rho^\prime(x,y) = \min\{1,\rho(x,y)\}$ is a metric [duplicate]

Let $(X,\rho)$ be a metric space. Define $\rho^\prime: X \times X \to \mathbf{R}$ by $\rho^\prime (x,y) = \min\{1,\rho(x,y)\}$ for all $x, y \in X$. Does anyone know of a short proof that $\rho^\prime$ satisfies the triangle inequality? It is easy (but tedious) to verify by checking cases.

For $x,y,z \in X$, we have $$ \rho'(x,z) = \\ \min\{1,\rho(x,z)\} \leq\\ \min\{1,\rho(x,y) + \rho(y,z)\} \leq\\ \min\{1,\rho(x,y)\} + \min\{1,\rho(y,z)\} =\\ \rho'(x,y) + \rho'(y,z) $$ In order to make this proof work, it suffices to show that the two inequalities used actually hold. In particular, we need to show (or perhaps simply state) the following:

• For the first: if $a \leq b$, then $\min\{1,a\} \leq \min\{1,b\}$
• For the second: for any non-negative real $a,b$, $\min\{1,a+b\} \leq \min\{1,a\} + \min\{1,b\}$

The first statement can be checked in two cases. The second might require $3$.

I think you'll find it easier to prove these statements separately than to separate the whole proof into cases.

If you don't want to check different cases, keep in mind that $\min(a,b) = \frac{a+b-|a-b|}{2}$, it might help direct calculations.

By the way, even if it is not relevant here, the same trick exists for $\max(a,b) = \frac{|a-b|+a+b}{2}$.