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How can I prove the following inequality:

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

Thank you.

  • 3
    Note: the answer is *highly* dependent on how much you already know about inequalities.2012-03-21

2 Answers 2

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

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...

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    Oh, right. Awesome, thank you :-)2012-03-21
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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 $a^2>\sqrt{b}$.

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    @Anonymous By definition, according to any definition I've seen.2012-03-21