Inequality

teleport

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

If we know that [tex](\frac{a - 1}{1 + a})^{n + 1} \geq \displaystyle\prod_{i=0}^n\frac{x_i - 1}{1+x_i}[/tex] is the inequality
[tex] a^{n+1} \geq \displaystyle\prod_{i=0}^n\ x_i [/tex] 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:

[tex] 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 [/tex].

Now it would be great if I could assume that if it works for n = 0 then
[tex] a \geq x_0 [/tex] and therefore [tex] a \geq x [/tex] for all n since I can allways permute the highest of the x and set it as [tex] x_0[/tex]. 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. Im trying to fix some

teleport

Can it be proven here that [tex] a \geq x [/tex] for all n, or should I just say that the inequality is true iff [tex] a \geq x [/tex] for all n?

teleport

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

[tex](\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} [/tex].

But since

[tex] f(x) = \frac{x - 1}{1 + x}[/tex]

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!