## Inequality

1. The problem statement, all variables and given/known data

If we know that $$(\frac{a - 1}{1 + a})^{n + 1} \geq \displaystyle\prod_{i=0}^n\frac{x_i - 1}{1+x_i}$$ is the inequality
$$a^{n+1} \geq \displaystyle\prod_{i=0}^n\ x_i$$ true? Prove your answer.

2. Relevant equations

Not sure

3. The attempt at a solution

I tried induction:

The base n = 0 works.

Assume it works for n -1

Proving it works for n:

$$a^{n +1} = aa^n \geq a\displaystyle\prod_{i=0}^{n - 1} x_i = \frac{a}{x_n}\displaystyle\prod_{i=0}^{n} x_i$$.

Now it would be great if I could assume that if it works for n = 0 then
$$a \geq x_0$$ and therefore $$a \geq x$$ for all n since I can allways permute the highest of the x and set it as $$x_0$$. If this is true, then I would get the result immediately. But I don't really know if I could do this. Any help is appreciated. I am very interested to see if the inequality could be proven without induction. Thanks for any comments.
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 wow my latex sucks. sorry for that. Im trying to fix some
 Can it be proven here that $$a \geq x$$ for all n, or should I just say that the inequality is true iff $$a \geq x$$ for all n?

## Inequality

Oh I got it now. I didn't realize that

$$(\frac{a - 1}{a + 1})^{n + 1} = \displaystyle\prod_{i=0}^n\ (\frac{a - 1}{a + 1}) \geq \displaystyle\prod_{i=0}^n\ \frac{x - 1}{1 + x}$$.

But since

$$f(x) = \frac{x - 1}{1 + x}$$

is an increasing function, then a must be greater than x for all n. Is this reasoning correct? I am suspicious of that. I need some confirmation. Thanks
 Wait, no I don't think that's true. Man!