Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove the following inequality:

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

Thank you.

share|improve this question
3  
Note: the answer is highly dependent on how much you already know about inequalities. –  Alex Becker Mar 21 '12 at 16:25
add comment

2 Answers

up vote 18 down vote accepted

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

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...

share|improve this answer
    
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
1  
Also, (+1) for interaction beyond just asking the question. –  The Chaz 2.0 Mar 21 '12 at 16:41
1  
Oh, right. Awesome, thank you :-) –  Anonymous Mar 21 '12 at 16:42
add comment

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 $a^2>\sqrt{b}$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.