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I've been testing this with many values and it seems to always be true. I've been trying to rework the inequality into a form where it's much more obvious that the left hand side is always less than the right, but can't seem to do it. Can anyone help me out here?

Thanks.

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    $\begingroup$ Generalise to arbitrary numbers: $2ab \leqslant a^2 + b^2$. $\endgroup$ – Daniel Fischer Oct 9 '14 at 18:57
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    $\begingroup$ square both sides $\endgroup$ – John Oct 9 '14 at 18:58
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$$ 2\sqrt{n}\sqrt{n+1}=\sqrt{4n^2+4n}<\sqrt{4n^2+4n+1}=2n+1. $$ Note. This inequality holds for every non-negative real $n$ (not only integer.)

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  • $\begingroup$ Thanks, that's exactly what I was looking for! $\endgroup$ – user176049 Oct 9 '14 at 19:00
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    $\begingroup$ It's worth noting that this proves it for all $n\ge0$ not just integers $\endgroup$ – Alice Ryhl Oct 9 '14 at 19:01
  • $\begingroup$ That is in fact right! $\endgroup$ – Yiorgos S. Smyrlis Oct 9 '14 at 19:01
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By the AM-GM inequality, $\displaystyle\sqrt{n}\sqrt{n+1}\le\frac{2n+1}{2}$.

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  • $\begingroup$ This should be a standard tool in any inequality toolbox. $\endgroup$ – Steven Stadnicki Oct 13 '14 at 21:58
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$$2n+1 - 2\sqrt n\sqrt{n+1}=(\underbrace{\sqrt n-\sqrt{n+1}}_{\ne 0})^2 >0$$

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