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An interesting little fact I noticed today while problem-solving:

Show that, for any positive reals $x_1, x_2, ... x_k$ and any positive integers $m, n$ with $n >m$,

$$ x_1 x_2 \cdots x_k \ge 1 \implies {x_1}^m + {x_2}^m + \dots + {x_k}^m \le {x_1}^n + {x_2}^n + \dots + {x_k}^n.$$

Hint: Start with $m = 1$, $n = 2$, and apply Chebyshev. Generalize!

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Suppose that $x_1\leq x_2\leq \dots \leq x_k$: $$ x_1 x_2 \cdots x_k \ge 1 \implies \\ {x_1}^m + {x_2}^m + \dots + {x_k}^m \le ({x_1}^m + {x_2}^m + \dots + {x_k}^m)(x_1x_2\dots x_k)^{\frac{m-n}{k}}\\ \leq ({x_1}^m + {x_2}^m + \dots + {x_k}^m)\left(\frac{x_1^{m-n}+x_2^{m-n}+\dots+ x_k^{m-n}}{k}\right)\\ $$ Now we can use Chebyshev inequality for the last step and we get: $$ \left(\frac{{x_1}^m + {x_2}^m + \dots + {x_k}^m}{k}\right)\left(\frac{x_1^{m-n}+x_2^{m-n}+\dots+ x_k^{m-n}}{k}\right)\leq \frac{{x_1}^n + {x_2}^n + \dots + {x_k}^n}{k} $$ which proves the desired inequality.

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  • $\begingroup$ +1 - While this was not my intended solution, it is extremely clever. $\endgroup$ – Chris Oct 19 '13 at 21:41

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