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I am stuck on proving the following inequality:

Let: $x_{1},x_{2},...,x_{n}\geq 0$. Prove that:$(x_{1}+x_{2}+...+x_{n})\leq n^{n} x_{1}x_{2}...x_{n}$

where $n$ is a natural number.

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    @J.D: It's not entirely incorrect (mathematically). However I would like to 'improve' the title (in suggestion) to the current version on the ground of 'changing the tone'. "Prove the inequality" seems a bit commanding to me, which (in my opinion) somewhat alters OP's tone. That said, I should had hit the 'approve' button instead of 'reject' button, my apologies.2012-03-10

3 Answers 3

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For $n=2$, set $x_1=\frac 1 2$ and $x_2=\frac 1 4$ to see that, if your inequality held true, $\frac 3 4 \le 4\cdot\frac 1 2 \cdot \frac 1 4=\frac 1 2$ which is a contradiction.


So, you must have claimed that, $(x_1+x_2 \cdots +x_n)^n \ge n^n (x_1 x_2 \cdots x_n)$ for positive values of $x_i$ and for all $n \in \Bbb N$.

This nicely rearranges to give,

$\dfrac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}$

Note that this is nothing but the assertion that $AM \ge GM$ where $AM$ stands for the arithmetic mean and $GM$ stands for the geometric mean.

Note that for $n=2$, the proof of this inequality relies on the fact that square numbers are positive. Instead of my reproducing a proof here, I suggest you'd look into Wikipedia link.

Corollary:

If the sum $x_1+x_2+x_3+ \cdots + x_n \le 1$, we have that $x_1+x_2+ \cdots +x_n \ge (x_1+x_2+x_3+\cdots+x_n)^n \ge n^n x_1x_2\cdots x_n \\ \quad \\ \boxed{x_1+x_2+\cdots + x_n \ge n^n x_1x_2 \cdots x_n}$

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    @PeterT.off Sure. =) (But I am too lazy now to fix it.)2012-03-10
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The inequality of the post, as currently stated, is not universally correct.

It fails immediately if some but not all the $x_i$ are $0$. But it also fails when all the $x_i$ are positive. The failure is essentially automatic, by continuity, or because the inequality is not homogeneous.

Let $\displaystyle x_i=\frac{1}{n^2}$ for all $n$.

Then $\displaystyle\sum x_i=\frac{1}{n}$.

But $\displaystyle\prod x_i=\frac{1}{n^{2n}}$, so $\displaystyle n^n\prod x_i=\frac{1}{n^n}$.

The right-hand side, when $n>1$, is less than the left-hand side.

If the inequality is to hold, conditions other than positivity must be put on the $x_i$.

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    Make that homogeneous symmetric polynomials...2012-03-10
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False as said by others, an easy counterexample is reached with $x_1=77$ and $x_k=0$ for $k=2,\ldots n$.