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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? –  user21436 Mar 10 '12 at 3:42
    
@Kannappan Sampath: and $\ge$ instead of $\le$ ?! –  Quixotic Mar 10 '12 at 3:43
    
Hah! I missed that too. Thanks for pointing that out. : ) –  user21436 Mar 10 '12 at 3:44
    
@S.k.J: This is related: Inequality of arithmetic and geometric means –  Quixotic Mar 10 '12 at 3:48
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With no restrictions other than non-negativity, the actual result is $$ (x_1+x_2+x_3+\cdots+x_n)^n \ge n^n x_1x_2\cdots x_n $$ –  Will Jagy Mar 10 '12 at 5:03
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3 Answers

up vote 5 down vote accepted

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 $x_{1} = x_{2} = 10$? –  Chris Leary Mar 10 '12 at 3:54
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@ChrisLeary The inequality OP claims does hold. : ) But, I don't quite understand the point you're trying to make. –  user21436 Mar 10 '12 at 4:01
    
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. –  Chris Leary Mar 10 '12 at 4:28
    
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. –  JohnJamesSmith Mar 10 '12 at 4:35
    
@KannappanSampath ".. nothing but the assertion that $AM \geq GM$"? –  Pedro Tamaroff Mar 10 '12 at 4:43
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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 $$ (x_1+x_2+x_3+\cdots+x_n)^n \ge n^n x_1x_2\cdots x_n $$ –  Will Jagy Mar 10 '12 at 5:03
    
@Will Jagy: That fix certainly works, the makeover turns it into AM-GM. –  André Nicolas Mar 10 '12 at 5:12
    
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 $t \rightarrow \infty.$ –  Will Jagy Mar 10 '12 at 5:52
    
Make that homogeneous symmetric polynomials... –  Will Jagy Mar 10 '12 at 6:48
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False as said by others, an easy counterexample is reached with $x_1=77$ and $x_k=0$ for $k=2,\ldots n$.

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