We can show that hence we have either or . If it is the second case, put for and compute
Let x1,x2,...,xn be positive numbers satisfying
Show that for any t>=0,
Nice. Could you explain one thing? If I understand correctly, you claim that , from where either the first or the second factor is . Theoretically, the choice of which factor exceeds 1 can depend on . However, it seems that later you require that either
for all or for all . This allows you substituting for in case the second alternative is true. But is it possible, say, that the second alternative is true for , but the first alternative is true for ?
I think we can show the result by induction. It's not difficult for and it's true for and were is positive.
We assume and (by symmetry ). Let if and . We have for any hence . Now we have to show that
. If we compute the difference we get which is positive.