# Inequality proof by induction

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?

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

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.

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.