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Suppose $x_1, \cdots, x_n$ are strictly positive real numbers. Then

$(x_1 + \cdots +x_n)^n \geq n(x_1 \cdots x_n)^{1/n}$

Is this statement true? If not, can you provide a counter example?

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    $\begingroup$ I think the statement is FALSE.. $\endgroup$
    – Empty
    Commented Jul 22, 2015 at 15:06
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    $\begingroup$ Take, $x_1=1,x_2=2$. Then $\sqrt 3>2\sqrt 2$ , which is absurd. $\endgroup$
    – Empty
    Commented Jul 22, 2015 at 15:07
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    $\begingroup$ It is false as the counter examples peole gave you prove. But the inequality $x_1+\cdots+x_n\ge n(x_1\cdots x_n)^\frac{1}{n}$ is correct. It's the AM-GM inequality (Inequality of arithmetic and geometric means). $\endgroup$ Commented Jul 22, 2015 at 15:10

3 Answers 3

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Just consider e.g. $n=2$ and $x_1 = x_2 = \frac{1}{4}$.

Note that the AM-GM inequality is of a very similar form; it states that

$$x_1+\ldots+x_n \geq n (x_1 \cdots x_n)^{\frac{1}{n}}.$$

This means in particular that the inequality from the OP holds true whenever $x_1+\ldots+x_n \geq 1$.

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Put $x_1=x_2=\cdots=x_n=x$. The inequality becomes $(nx)^{n-1}\ge1$, which is not generally true.

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It's true, when $x_1+\cdots+x_n>1$, because in that case, it follows from what is known as the Inequality of arithmetic and geometric means, which states that

$$x_1 + \cdots +x_n \geq n(x_1 \cdots x_n)^{1/n},$$

and when $\sum x_i > 1$, you have $(x_1 + \cdots +x_n)^n \geq x_1 + \cdots +x_n$.

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