7
$\begingroup$

How can I prove the following inequality:

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

Thank you.

$\endgroup$

closed as off-topic by user21820, GNUSupporter 8964民主女神 地下教會, Saad, José Carlos Santos, Cesareo Oct 8 '18 at 16:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, GNUSupporter 8964民主女神 地下教會, Saad, José Carlos Santos, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Note: the answer is highly dependent on how much you already know about inequalities. $\endgroup$ – Alex Becker Mar 21 '12 at 16:25
21
$\begingroup$

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

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...

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – Anonymous Mar 21 '12 at 16:38
  • $\begingroup$ $a$ 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. $\endgroup$ – The Chaz 2.0 Mar 21 '12 at 16:41
  • 1
    $\begingroup$ Also, (+1) for interaction beyond just asking the question. $\endgroup$ – The Chaz 2.0 Mar 21 '12 at 16:41
  • 1
    $\begingroup$ Oh, right. Awesome, thank you :-) $\endgroup$ – Anonymous Mar 21 '12 at 16:42
6
$\begingroup$

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

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

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