# How to show that $\min\{\alpha,\max\{\beta,\gamma\}\}=\max \{ \min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}.$

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.

• I wouldn't consider $6$ to be "too many cases", but you can cut down to $3$ cases because of the symmetry between $\beta$ and $\gamma$. – Andreas Blass Mar 24 '18 at 14:03
• You can cut it down to two cases: Either $\alpha\ge \max(\beta,\gamma)$ or $\alpha<\max(\beta,\gamma)$ – Mastrem Mar 24 '18 at 14:11

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: $$\max(\min(\alpha,\beta),\min(\alpha,\gamma))=\max(\alpha,\min(\alpha,\min(\beta,\gamma)))=\max(\alpha,\min(\alpha,\beta,\gamma))=\alpha=\min(\alpha,\max(\beta,\gamma))$$
First case: $\alpha=\max\{\beta,\gamma\}$ then $$\min\{\alpha,\max\{\beta,\gamma\}\}=\alpha$$ and $$\max\{\min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}=\max\{\beta,\gamma\}=\alpha$$
Second case: $\alpha>\max\{\beta,\gamma\}$ then $\alpha>\beta$ and $\alpha>\gamma$. These imply $$\min\{\alpha,\max\{\beta,\gamma\}\}=\max\{\beta,\gamma\}$$ and $$\max\{\min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}=\max\{\beta,\gamma\}$$
Third case (last one): $\alpha<\max\{\beta,\gamma\}$ then $\alpha<\beta$ or $\alpha<\gamma$. Therefore $$\min\{\alpha,\max\{\beta,\gamma\}\}=\alpha$$ and $$\max\{\min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}=\max\{\alpha,\min\{\alpha,\gamma\}\}=\alpha$$ or $$\max\{\min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}=\max\{\min\{\alpha,\beta\},\alpha\}=\alpha$$ In either case the answers equal.