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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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    Are you missing an exponent $n$ in LHS?2012-03-10
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    @Kannappan Sampath: and $\ge$ instead of $\le$ ?!2012-03-10
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    Hah! I missed that too. Thanks for pointing that out. : )2012-03-10
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    @S.k.J: This is related: [Inequality of arithmetic and geometric means](http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means)2012-03-10
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    With no restrictions other than non-negativity, the actual result is $$ (x_1+x_2+x_3+\cdots+x_n)^n \ge n^n x_1x_2\cdots x_n $$2012-03-10
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    @Foool how is this [edit suggestion](http://math.stackexchange.com/suggested-edits/5093) "incorrect or an attempt to reply to or comment on the existing post"?2012-03-10
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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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    But what happens if $n=2$ and $x_{1} = x_{2} = 10$?2012-03-10
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    @ChrisLeary The inequality OP claims does hold. : ) But, I don't quite understand the point you're trying to make.2012-03-10
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    The original question did not place any restriction on the $x_{i}$ other than that they be nonnegative. My observation was that the OP's inequality cannot hold without restriction on the $x_{i}$. If their sum does not exceed $1$, the inequality follows per your corollary.2012-03-10
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    I do not see how the $\sum x_i \geq n^n \prod x_i$ reduces to the AM-GM inequality. Chris Leary's comment is a counter-example, in fact: the statement that $10 + 10 \geq 2^2 \cdot 10 \cdot 10$ is false. I think I must be missing something.2012-03-10
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    @KannappanSampath ".. **nothing but** the assertion that $AM \geq GM$"?2012-03-10
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    With no restrictions other than non-negativity, the actual result is $$ (x_1+x_2+x_3+\cdots+x_n)^n \ge n^n x_1x_2\cdots x_n $$2012-03-10
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    People, have I missed something, other than the exponent that was fixed by anon. Then, why the downvote? Will the downvoter clarify?2012-03-10
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    @PeterT.off I don't understand your comment at all. I'd like it if you expanded it a bit more.2012-03-10
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    @JohnJamesSmith Did you read the condition that is in bold? Your "counter-example" does not satisfy that condition and is not even a counter example therefore.2012-03-10
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    @ChrisLeary Is your comment sort of not what I had observed already through a corollary and counter example?2012-03-10
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    @KannappanSampath I didn't downvote or anything. I just thing the AM-GM ineq. deserves more respect. =)2012-03-10
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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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    With no restrictions other than non-negativity, the actual result is $$ (x_1+x_2+x_3+\cdots+x_n)^n \ge n^n x_1x_2\cdots x_n $$2012-03-10
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    @Will Jagy: That fix certainly works, the makeover turns it into AM-GM.2012-03-10
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    Even if we keep the non-negativity restriction, the only possibility for an inequality comparing one symmetric polynomial to another is raising one side to an exponent that makes the degrees become equal. Without that, whatever holds for some fixed point $\vec{p},$ that cannot hold for $t \vec{p}$ for all $t > 0.$ Opposite results as $t \rightarrow 0$ and as $t \rightarrow \infty.$2012-03-10
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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 $k=2,\ldots n$.