How can I prove the following inequality:

Given $ a,b>0 $ and $a^2>b $, we have $a>\sqrt b$

Thank you.


closed as off-topic by user21820, GNUSupporter 8964民主女神 地下教會, Saad, José Carlos Santos, Cesareo Oct 8 '18 at 16:33

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    $\begingroup$ Note: the answer is highly dependent on how much you already know about inequalities. $\endgroup$ – Alex Becker Mar 21 '12 at 16:25

$a^2 > b \Leftrightarrow (a - \sqrt{b})(a + \sqrt{b}) > 0$

Both of these factors must be positive, since both $a$ and $\sqrt{b}$ are positive. In particular, $a - \sqrt{b} > 0$

Indeed, I stand on the shoulders of giants...

  • $\begingroup$ Both of these factors multiplied with each other gives us positive- but why can't each of this factor be negative?(negative multiply negative gives us positive) $\endgroup$ – Anonymous Mar 21 '12 at 16:38
  • $\begingroup$ $a$ is positive. $\sqrt{b}$ is positive. When you add them, you get the positive number $ a + \sqrt{b}$, which is the second factor. So the first factor must also be positive. $\endgroup$ – The Chaz 2.0 Mar 21 '12 at 16:41
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    $\begingroup$ Also, (+1) for interaction beyond just asking the question. $\endgroup$ – The Chaz 2.0 Mar 21 '12 at 16:41
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    $\begingroup$ Oh, right. Awesome, thank you :-) $\endgroup$ – Anonymous Mar 21 '12 at 16:42

Suppose otherwise, i.e. that $a\leq \sqrt{b}$. Then $a^2=a\cdot a\leq \sqrt{b}\cdot a\leq \sqrt{b}\cdot\sqrt{b}=b$, so $a^2\leq b$, contradicting the fact that $a^2>\sqrt{b}$.

  • $\begingroup$ Awesome, thank you! :-) $\endgroup$ – Anonymous Mar 21 '12 at 16:57
  • $\begingroup$ BTW, is $\sqrt{b}\cdot\sqrt{b}=b$ by definition or can it be proven? $\endgroup$ – Anonymous Mar 21 '12 at 17:00
  • $\begingroup$ @Anonymous By definition, according to any definition I've seen. $\endgroup$ – Alex Becker Mar 21 '12 at 23:49

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