Is this inequality proof possible without induction?

[teleport]

Homework Statement

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.

Homework Equations

Not sure

The Attempt at a Solution

I tried induction:

The base n = 0 works.

Assume it works for n -1

Proving it works for n:

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

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.

 

[teleport]

wow my latex sucks. sorry for that. I am trying to fix some

 

[teleport]

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?

 

[teleport]

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

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

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

 

[teleport]

Wait, no I don't think that's true. Man!