Prove that $\max(n,\min(k,q))=\min(\max(n,k),\max(n,q))$.

36 Views Asked by At

Prove that $\max(n,\min(k,q))=\min(\max(n,k),\max(n,q))$ where $n,k,q\in\Bbb R$.

To prove this I tested what happens in each one of the $6$ permutations of inequalities, i.e. $n<k<q$ and $q<k<n$ for example. In each of the $6$ order permutations it holds true, does that mean it is proven?

1

There are 1 best solutions below

0
On

Possibly not. Have you tested any of the equality cases?

Also note that this is symmetrical in k and q, So once you have proved it for one ordering of k and q, the result is also true for the other ordering.