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
(x1)(x2)(x3)...(xn)=1.
Show that for any t>=0,
(1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.
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 ?
You're right, what I did cannot be correct because I didn't use the fact that .
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.