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How do you show that if $x_1,\ldots,x_n$ are real numbers, then $|x_1|\ldots|x_n| \le |x_1|^2 + \cdots + |x_n|^2$. I tried using induction, but I'm guessing that's not the way to do it? Help would be appreciated.

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What do the dots on the left-hand side mean? Is it supposed to be a product? Or something else? In any case, take $n = 1$ and $x_1 = 1/2$ for a counterexample. – cardinal May 31 '12 at 23:07
Maybe you can prove it with an extra $n$ in there... – Pedro Tamaroff May 31 '12 at 23:11
Have you considered accept answers? You have asked 7 questions, all with at least one answer! – leo Jun 1 '12 at 0:15
@Michael I don't know that \cdots is the right edit for the OP's three periods on the left. \cdots implies multiplication, and it's unclear that's what OP wants. Maybe \ldots instead? – alex.jordan Jun 1 '12 at 0:23
The right kind of such inequalities should be homogeneous of the same degree on both sides. Here, it is homogeneous of degree $n$ on the left and $2$ on the right. – Francesco Sica Jun 1 '12 at 3:13

You can't.

Let $x_k = 2$. Then $|x_1|\cdots|x_n| = 2^n$, but $|x_1|^2 + \cdots + |x_n|^2 = 4n$. Choose $n=5$, then $32>20$.

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Maybe this would help to clear the original statement:

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