# prove srt(ab) => (a+b)/2

• May 1st 2013, 02:12 PM
kingsolomonsgrave
prove 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 PM
Plato
Re: prove srt(ab) => (a+b)/2
Quote:

Originally Posted by kingsolomonsgrave
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

You have it written incorrectly.

$\displaystyle \\(x-y)^2=x^2-2xy+y^2\ge 0\\x^2+y^2\ge 2xy$

If $\displaystyle a\ge 0~\&~b\ge 0$ then let $\displaystyle x=\sqrt a~\&~b=\sqrt b$.
• May 1st 2013, 05:38 PM
Prove It
Re: prove srt(ab) => (a+b)/2
Quote:

Originally Posted by kingsolomonsgrave
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

Are you sure your original equation wasn't \displaystyle \displaystyle \begin{align*} \sqrt{ab} \leq \frac{a + b}{2} \end{align*}?
• May 2nd 2013, 04:31 AM
kingsolomonsgrave
Re: 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 AM
Prove It
Re: prove srt(ab) => (a+b)/2
Quote:

Originally Posted by kingsolomonsgrave
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.

\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 AM
abualabed
Re: prove srt(ab) => (a+b)/2
geometrically
• May 3rd 2013, 08:52 AM
kingsolomonsgrave
Re: prove srt(ab) => (a+b)/2
How do we know the vertical line is sqrt(ab)?
• May 3rd 2013, 09:57 AM
emakarov
Re: prove srt(ab) => (a+b)/2
Quote:

Originally Posted by kingsolomonsgrave
How do we know the vertical line is sqrt(ab)?

That's because the two small right triangles into which the big triangle is split are similar.
• May 3rd 2013, 12:16 PM
kingsolomonsgrave
Re: 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 PM
emakarov
Re: prove srt(ab) => (a+b)/2
Quote:

Originally Posted by kingsolomonsgrave
How does that tell me that that length is $\displaystyle \sqrt{ab}$?

If the altitude of the big triangle is h, then it follows from the similarity of small triangles that a / h = h / b.
• May 3rd 2013, 12:44 PM
Plato
Re: prove srt(ab) => (a+b)/2
Quote:

Originally Posted by kingsolomonsgrave
Thanks! How does that tell me that that length is $\displaystyle \sqrt{ab}$?

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 .