Intereting Posts

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All groups where underlying set has at most 4 elements

How can I prove the following inequality:

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

Thank you.

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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…

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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