How can I prove the following inequality:
Given $ a,b>0 $ and $a^2>b $, we have $a>\sqrt b$
Thank you.
How can I prove the following inequality:
Given $ a,b>0 $ and $a^2>b $, we have $a>\sqrt b$
Thank you.
$a^2 > b \Leftrightarrow (a - \sqrt{b})(a + \sqrt{b}) > 0$
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 $a^2>\sqrt{b}$.