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Let $z_1,\dots,z_n$ be $n$ real numbers with $n \ge z_i \ge 1$ for all $i$. I can show that $$ \left(\sum_{i =1}^n (z_i^2-1)\right)^k \ge \sum_{i =1}^n (z_i^{2k}-1), $$ for all $k \ge 1$ by induction on $k$. But I wonder if there is a way to prove this directly without induction.

Note that if we remove the $-1$ this is just the lp-norm inequality.


NOTE: The inequality is not true as it's stated above. See the accepted answer below. But I do believe it should be true if we further assume $z_i$ are integers.

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The inequality does not hold if $n \ge 2$ and $k > 1$. As a counter-example, choose $z_1 = \ldots = z_n = \sqrt x > 1$. Then your inequality becomes $$ n^k(x-1)^k \ge n(x^k - 1) \, . $$ But the difference $$ f(x) = n^k(x-1)^k - n(x^k - 1) $$ satisfies $f(1) = 0$ and $f'(1) < 0$, so that $f(x) < 0$ at least in some interval $(1, 1 + \varepsilon)$.

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  • $\begingroup$ But I assumed that $z_i \ge 1$ for all $i$. $\endgroup$ – ablmf Jun 28 '17 at 4:50
  • $\begingroup$ @ablmf: You are right, my fault. – But then your claim is wrong at least for $n=1$. $\endgroup$ – Martin R Jun 28 '17 at 4:56
  • $\begingroup$ Yeah, sorry, I forgot to add $z_i \le n$. That's what I meant. $\endgroup$ – ablmf Jun 28 '17 at 5:32
  • $\begingroup$ @ablmf: It is still wrong, see update. $\endgroup$ – Martin R Jun 28 '17 at 5:58
  • $\begingroup$ you're right, this does not work for real numbers. I implicitly assumed that $z_i$'s are integers. Thanks for pointing out the mistake. $\endgroup$ – ablmf Jun 28 '17 at 6:46

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