how would I go about proving this?

$\displaystyle sqrt(ab) \geq (a+b)/2$

i manipulated the expression so that $\displaystyle 0 \geq (a-b)^2$ but im not sure if that proves anything

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- May 1st 2013, 02:12 PMkingsolomonsgraveprove srt(ab) => (a+b)/2
how would I go about proving this?

$\displaystyle sqrt(ab) \geq (a+b)/2$

i manipulated the expression so that $\displaystyle 0 \geq (a-b)^2$ but im not sure if that proves anything - May 1st 2013, 02:25 PMPlatoRe: prove srt(ab) => (a+b)/2
- May 1st 2013, 05:38 PMProve ItRe: prove srt(ab) => (a+b)/2
- May 2nd 2013, 04:31 AMkingsolomonsgraveRe: prove srt(ab) => (a+b)/2
yes! I knew something was off. It should have been:

it should have been:

Prove $\displaystyle \sqrt{ab}\leq \frac{a+b}{2}$ for all positive integers. - May 2nd 2013, 05:26 AMProve ItRe: prove srt(ab) => (a+b)/2
$\displaystyle \displaystyle \begin{align*} 0 &\leq \left( a - b \right) ^2 \\ 0 &\leq a^2 - 2ab + b^2 \\ 4ab &\leq a^2 + 2ab + b^2 \\ 4ab &\leq \left( a + b \right) ^2 \\ ab &\leq \frac{ \left( a + b \right) ^2}{4} \\ ab &\leq \left( \frac{a + b}{2} \right) ^2 \\ \sqrt{ab} &\leq \frac{a + b}{2} \end{align*}$

- May 2nd 2013, 08:23 AMabualabedRe: prove srt(ab) => (a+b)/2
geometrically

- May 3rd 2013, 08:52 AMkingsolomonsgraveRe: prove srt(ab) => (a+b)/2
How do we know the vertical line is sqrt(ab)?

- May 3rd 2013, 09:57 AMemakarovRe: prove srt(ab) => (a+b)/2
- May 3rd 2013, 12:16 PMkingsolomonsgraveRe: prove srt(ab) => (a+b)/2
Thanks! How does that tell me that that length is $\displaystyle \sqrt{ab}$?

I can see that the smaller triangle is proportional to the larger and that they have the same angles.

The most I can come up with is that if a = b then you get a big right triangle with angles of 45 on either side. That way $\displaystyle a^2+a^2= \sqrt{2}$ and $\displaystyle b^2+b^2 = \sqrt{2}$ and $\displaystyle a=b so a^2=ab=b^2$ so the line is $\displaystyle \sqrt{ab}$ - May 3rd 2013, 12:33 PMemakarovRe: prove srt(ab) => (a+b)/2
- May 3rd 2013, 12:44 PMPlatoRe: prove srt(ab) => (a+b)/2
Have you heard the term

*mean proportional*?

If $\displaystyle \frac{a}{h}=\frac{h}{b}>0 $ then $\displaystyle h $ is the mean proportional between $\displaystyle a~\&~b $.

In a right triangle the altitude from the right angle to the hypotenuse is the mean proportional between the parts.

That true in the supplied diagram .