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.

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- March 7th 2011, 05:35 AMchris86Proving (1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.
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. - March 7th 2011, 11:35 AMgirdav
We can show that hence we have either or . If it is the second case, put for and compute

- March 7th 2011, 12:44 PMemakarov
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 ? - March 8th 2011, 10:03 AMgirdav
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. - March 10th 2011, 04:30 PMawkward