Square Root Inequality [closed]

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, CesareoOct 8 '18 at 16:33

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• Note: the answer is highly dependent on how much you already know about inequalities. – 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...

• 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) – Anonymous Mar 21 '12 at 16:38
• $a$ 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. – The Chaz 2.0 Mar 21 '12 at 16:41
• Also, (+1) for interaction beyond just asking the question. – The Chaz 2.0 Mar 21 '12 at 16:41
• Oh, right. Awesome, thank you :-) – 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 $a^2>\sqrt{b}$.

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