# More with AM and GM

• Jun 6th 2006, 01:57 PM
Susie38
More with AM and GM
IF a,b >0
prove: AM - GM >= GM - HM
• Aug 26th 2006, 07:49 PM
Rebesques
If I get it right, u need to show that for all a,b>0,

$\displaystyle \frac{a+b}{2}-\sqrt{ab}\geq\sqrt{ab}-\frac{2}{\frac{1}{a}+\frac{1}{b}}$.

Both sides are positive; Begin calculating until you get to a valid inequality.
(remember AM>GM>HM.)
• Nov 21st 2006, 04:35 PM
putnam120
$\displaystyle \frac{a+b}{2}-\sqrt{ab}\ge \sqrt{ab}-\frac{2}{\frac{1}{a}+\frac{1}{b}}$

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

$\displaystyle \frac{a+b}{2}+\frac{2ab}{a+b}\ge 2\sqrt{ab}$

from here we just have to apply AM-GM. Note that $\displaystyle \left(\frac{a+b}{2}\right)\left(\frac{2ab}{a+b}\ri ght)=ab$.

if you don't know AM-GM, basically it just states: for any $\displaystyle x,y\ge 0$ we are assured that $\displaystyle x+y\ge 2\sqrt{xy}$, with equality iff (if and only if) $\displaystyle x=y$.

Challenge: Prove that for all nonnegative x and y that

$\displaystyle \sqrt{\frac{x^2+y^2}{2}}-\frac{x+y}{2} \ge \sqrt{xy}-\frac{2}{\frac{1}{x}+\frac{1}{y}}$

try not to use the fact that $\displaystyle \sqrt{\frac{x^2+y^2}{2}}\ge\frac{x+y}{2}$