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I'm supposed to prove that for any integer $n\geq 2$, if $x_{1},\ldots,x_{n}$ are real numbers in $]0,1[$, then

$(1-x_{1})\cdot\ldots\cdot(1-x_{n}) > 1 - x_{1} - \ldots - x_{n}$

I am trying the induction method so I first tried to find if it's true for n=2 : $(1-x_{1})(1-x_{2}) > 1- x_{1} - x_{2}$

$1-x_{1}-x_{2}+x_{1}x_{2} > 1-x_{1}-x_{2}$ and this is true because $x_{1}x_{2}$ is positive

for n+1 : $(1-x_{1})(1-x_{2})...(1-x_{n})(1-x_{n+1}) > 1-x_{1}-...-x_{n}-x_{n+1}$

I am stuck here.. what should I do next?

Edit : I then did

$(1-x_{1}-...-x_{n})(1-x_{n+1}) > 1-x_{1}-...-x_{n}-x_{n+1}$

How do I compare $(1-x_{n+1})$ and $-x_{n+1}$ to find out that the left side is bigger?

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Suppose it is true for some $n$ as you've shown.

Then $(1-x_{1})(1-x_{2})\cdots(1-x_{n})(1-x_{n+1}) > (1-x_{1}-\cdots-x_{n})(1-x_{n+1})$

$=(1-x_{1}-\cdots-x_{n})-x_{n+1}+x_{1}x_{n+1}+\cdots+x_{n}x_{n+1}>1-x_{1}-\cdots-x_{n}-x_{n+1}$

as $x_{i}x_{n+1} > 0$.

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To simplify this step, let $y = x_1+x_2+\dots +x_n$. Then

$$(1-x_1-\dots-x_n)(1-x_{n+1})=(1-y)(1-x_{n+1}) = 1-y-x_{n+1} + yx_{n+1} > 1-y- x_{n+1}$$

Finally substitute in $y$ and finish the proof.

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Assume $(1-x_1)(1-x_2)...(1-x_n) > 1 - x_1 -x_2 ... -x_n$ as the indutive hypothesis.

Then certainly $x_{n+1}x_1 + ... + x_{n+1}x_n > 0$ since each number is positive.

Thus $1- x_1 -... - x_n -x_{n+1} + x_{n+1}x_1 + ... +x_{n+1}x_n > 1 - x_1 ... - x_n - x_{n+1}$.

The LHS can be rephrased as follows:

$1- x_1 -... - x_n -x_{n+1} + x_{n+1}x_1 + ... +x_{n+1}x_n = (1- x_1...-x_n)(1 - x_{n+1})$, so that $(1...-x_n)(1-x_{n+1}) > 1 -x_1 -...-x_n - x_{n+1}$.

Now by the inductive assumption that $(1-x_1)(1-x_2)...(1-x_n) > 1 - x_1 -x_2 ... -x_n$, multiply both sides by $(1-x_{n+1})$ which is positive, to get

$(1-x_1)(1-x_2)...(1-x_n)(1-x_{n+1}) > (1 - x_1 -x_2 ... -x_n)(1-x_{n+1}) > 1 - x_1 ... - x_n - x_{n+1}$, and thus $(1-x_1)(1-x_2)...(1-x_n)(1-x_{n+1}) > 1 - x_1 ... - x_n - x_{n+1}$ as desired.

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