Arithmetric mean vs. Geometric mean

**Slazenger3** · February 9th 2010, 06:34 PM

Show that the geometric mean, radical(ab), is always less than or equal to the arithmetic mean, (a+b)/2.

**Soroban** · February 9th 2010, 06:55 PM

Hello, Slazenger3!

Show that the geometric mean, $\sqrt{ab}$ , is always less than or equal to the arithmetic mean, $\frac{a+b}{2}$

For any two real numbers, $a$ and $b$ : . $(a-b)^2 \:\geq\:0$

We have: . $a^2 - 2ab + b^2 \:\geq \:0$

Add $4ab$ to both sides: . $a^2 + 2ab + b^2 \:\geq\:4ab$

Then we have: / $(a+b)^2 \:\geq\:4ab$

Divide by 4: . $\frac{(a+b)^2}{4} \:\geq \:ab \quad\Rightarrow\quad \left(\frac{a+b}{2}\right)^2 \;\geq\;ab$

Take square root: . $\frac{a+b}{2} \:\geq\:\sqrt{ab}$

Therefore, the arithmetic mean is always greater than or equal to the geometric mean.

**Pulock2009** · February 9th 2010, 09:16 PM

infact for any n postive numbers, A.M.>=G.M.

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