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**ssff****Guest**

This is in the real numbers.

(1+a1)(a1+a2).....(1+an)>1+(a1+a2+...+an)

And n>=2, ai>0 (i=1...n).

Thanks.

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

I'm not sure if I understand what you're asking, but here goes.

I think here:

(1+a1)(a1+a2)

You meant to say: (1+a1)(1+a2)

...as that would make a lot more sense.

n >=2, so let's start with the minimum case: n=2.

(1+a1)(1+a2) = 1 + a1 + a2 + a1a2

1+(a1+a2) = 1 + a1 + a2

The first expression > the second, because there's that extra a1a2 term in there.

This will be true for any n >= 2! This is because the expanded multiplication will always have all the terms of the simple addition, plus some additionals that are combinations of the a's.

That proof probably wouldn't satisfy a mathematician, but it sure works for me.

El que pega primero pega dos veces.

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