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Prove that for $x_i > 0$ and $x_i$ distinct such that $x_1 < x_2... < x_{2n+1}$,

$\displaystyle\sum_{i = 1}^{2n + 1} (-1)^{i+1} x_i \leq \left(\sum_{i = 1}^{2n+1} (-1)^{i+1} (x_i)^n\right)^{\frac{1}{n}}$.

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    If we have $n=1$ and $x_1=2$ and $x_2=x_3=1$ we have on the left $1\cdot 2 -1+1$ and on the right side $(1\cdot 2 -1+1)^{\frac{1}{2}}$ which is less than the right hand side. If you take a closer look your equation says $f(n,x) \leq f(n,x)^{\frac1n} $ which does not hold in general2012-12-23
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    Sorry. $x_i$ should be ordered.2012-12-23
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    Induction? Yes?2012-12-23
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    I tried that but I got stuck.2012-12-23
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    You might have to formulate your induction hypothesis in a clever way-- see what different forms it can take on.2012-12-23

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To use induction, it is more convenient to prove the following more general inequality:

$$\left(\sum_{i = 1}^{2n + 1} (-1)^{i+1} x_i\right)^m \leq\sum_{i = 1}^{2n+1} (-1)^{i+1} x_i^m. \tag{1}$$ The original inequality is just the a special case of $(1)$ when $m=n$.

Let us fix $m$ and prove $(1)$ inductively on $n$. When $n=0$, $(1)$ is trivially true. Denote $$a=\sum_{i = 1}^{2n -1} (-1)^{i+1}x_i,\quad b=x_{2n+1}-x_{2n}\quad and\quad c=x_{2n}.$$ By induction, assume that $$a^m\le \sum_{i = 1}^{2n-1} (-1)^{i+1} x_i^m.$$ Then to obtain $(1)$, it suffices to prove that $$(a+b)^m+c^m\le a^m+(b+c)^m.\tag{2}$$

Noting that $a=x_{2n-1}-\sum_{i=1}^{n-1}(x_{2i}-x_{2i-1})

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    I'm very sorry. I didn't see that I made an error in the question.2012-12-23
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    @Victor: Never mind. I have updated my answer.2012-12-23