Show that the geometric mean, radical(ab), is always less than or equal to the arithmetic mean, (a+b)/2.
Hello, Slazenger3!
Show that the geometric mean, $\displaystyle \sqrt{ab}$, is always less than or equal to the arithmetic mean, $\displaystyle \frac{a+b}{2}$
For any two real numbers, $\displaystyle a$ and $\displaystyle b$: .$\displaystyle (a-b)^2 \:\geq\:0$
We have: .$\displaystyle a^2 - 2ab + b^2 \:\geq \:0$
Add $\displaystyle 4ab$ to both sides: .$\displaystyle a^2 + 2ab + b^2 \:\geq\:4ab $
Then we have: /$\displaystyle (a+b)^2 \:\geq\:4ab$
Divide by 4: .$\displaystyle \frac{(a+b)^2}{4} \:\geq \:ab \quad\Rightarrow\quad \left(\frac{a+b}{2}\right)^2 \;\geq\;ab $
Take square root: .$\displaystyle \frac{a+b}{2} \:\geq\:\sqrt{ab}$
Therefore, the arithmetic mean is always greater than or equal to the geometric mean.