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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.2012-05-31
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    Maybe you can prove it with an extra $n$ in there...2012-05-31
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    Have you considered accept answers? You have asked 7 questions, all with at least one answer!2012-06-01
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    @Michael: Your edits potentially change the interpretation of the question, which is unclear to begin with. Would it not be better to have the OP clarify the meaning of their post and notation?2012-06-01
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    Maybe the exponents on the right are supposed to be $n$'s? In that case the right side is at least as big as $X^n$, where $X=\max\{|x_1|,\ldots,|x_n|\}$. And then $X^n$ is at least as large as the left side, assuming the left side is a product.2012-06-01
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    @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?2012-06-01
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    If multiplication was not intended, then the OP should clarify. But for now I've changed it to \ldots.2012-06-01
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    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.2012-06-01

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