How to show that $\min\{\alpha,\max\{\beta,\gamma\}\}=\max \{ \min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}$ without considering too many cases? Here $\alpha,\beta,\gamma\in \mathbb{N}.$
I could consider $\alpha\leq \beta\leq \gamma$ and then various permutations of this inequality, but I was wondering if there was a cleverer argument.
PS. To give some context, this problem comes from the following problem: Let $a,b,c\in \mathbb{N}$ then show that $$a\wedge (b \vee c)=(a\wedge b)\vee (a\wedge c) $$ where $a\wedge b=\gcd(a,b)$ and $a\vee b=\text{lcm}(a,b)$ for any $a,b\in \mathbb{N}.$ Using prime factorizations for $a,b$ and $c$ one can reduce the problem to showing the equality that I have mentioned above.
Suppose that $\alpha\ge\max(\beta,\gamma)$, so $\alpha\ge\beta$ and $\alpha\ge\gamma$, meaning that $$\max(\min(\alpha,\beta),\min(\alpha,\gamma))=\max(\beta,\gamma)=\min(\alpha,\max(\beta,\gamma))$$ Otherwise, $\alpha<\max(\beta,\gamma)$, so: $$\max(\min(\alpha,\beta),\min(\alpha,\gamma))=\max(\alpha,\min(\alpha,\min(\beta,\gamma)))=\max(\alpha,\min(\alpha,\beta,\gamma))=\alpha=\min(\alpha,\max(\beta,\gamma))$$