# Thread: request help with proof

1. ## request help with proof

Show that (a + b)/2 is greater than or equal to squareroot (ab) for 0 < a ≤ b.

I'm in a class in which I'm being exposed to proofs for the first time, and I'm quite unsure about where to begin.

I can input examples to show that it's true, but that's obviously not what I need to do here.

While I would appreciate a complete explanation, even a gentle nudge in the right direction would be helpful.

Thanks!

2. Originally Posted by thejollywag
Show that (a + b)/2 is greater than or equal to squareroot (ab) for 0 < a ≤ b.

I'm in a class in which I'm being exposed to proofs for the first time, and I'm quite unsure about where to begin.

I can input examples to show that it's true, but that's obviously not what I need to do here.

While I would appreciate a complete explanation, even a gentle nudge in the right direction would be helpful.

Thanks!
haven't done this in awhile, but if I remember correctly, a proof may look somewhat like this

use the general idea that if $x \ge y$ , then $x - y \ge 0$

using the two values in question ...

$\frac{a+b}{2} - \sqrt{ab} =$

$\frac{a+b}{2} - \frac{2\sqrt{ab}}{2} =$

$\frac{a - 2\sqrt{ab} + b}{2} =$

$\frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0$

3. I appreciate it, Skeeter. That's an extremely helpful explanation.