Proof of Inequality of Square Roots of Two Reals

[turophile]

I posted this on the Calculus forum, but it's really a pre-calculus problem. I'm still looking for a hint, so I thought I would post it again here.

Prove: If a > b > 0, then sqrt(a) > sqrt(b).

Here's how I've started out:

a > b > 0 ==> sqrt(a)*sqrt(a) > sqrt(b)*sqrt(b) ==> sqrt(a) > b/sqrt(a) = (sqrt(a)*b)/a
a > b > 0 ==> sqrt(a)*sqrt(a) > sqrt(b)*sqrt(b) ==> a/sqrt(b) = (a*sqrt(b))/b > sqrt(b)

So I've got the two ends of the inequality, but I'm not sure how to link up the middle. Any hints?

 

[galactus]

Try this:

If $\displaystyle a>b>0$, then $\displaystyle a-b>0$.


or, equivalently, $\displaystyle (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})>0$

Divide both sides by $\displaystyle \sqrt{a}+\sqrt{b}$ and we get:

$\displaystyle \sqrt{a}-\sqrt{b}>0$

$\displaystyle \sqrt{a}>\sqrt{b}$

 

[BigGlenntheHeavy]

$\displaystyle Another \ way, \ proof \ by \ contradiction.$

$\displaystyle We \ have \ if \ a \ > \ b \ > \ 0, \ then \ prove \ \sqrt a \ > \ \sqrt b.$

$\displaystyle Suppose \ not, \ suppose \ \sqrt a \ \le \ \sqrt b, \ then \ \sqrt a \ = \ \sqrt b \ or \ \sqrt a \ < \ \sqrt b.$

$\displaystyle Case \ 1:$

$\displaystyle \sqrt a \ = \ \sqrt b \ \implies \ \sqrt a\sqrt a \ = \ \sqrt a\sqrt b \ = \ \sqrt b \ \sqrt b$

$\displaystyle Hence, \ a \ = \ b, \ but \ a \ > \ b, \ a \ contradiction.$

$\displaystyle Case \ 2:$

$\displaystyle \sqrt a \ < \ \sqrt b \ \implies \ \sqrt b \ > \ \sqrt a \ \implies \ \sqrt b\sqrt b \ > \ \sqrt a \ \sqrt a$

$\displaystyle hence, \ b \ > \ a, \ but \ b \ < \ a, \ another \ contradiction.$

$\displaystyle Ergo, \ the \ supposition \ is \ false, \ so \ a \ > \ b \ > \ 0 \ \implies \ \sqrt a \ > \ \sqrt b \ QED.$