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
$\displaystyle \displaystyle \begin{align*} 0 &\leq \left( a - b \right) ^2 \\ 0 &\leq a^2 - 2ab + b^2 \\ 4ab &\leq a^2 + 2ab + b^2 \\ 4ab &\leq \left( a + b \right) ^2 \\ ab &\leq \frac{ \left( a + b \right) ^2}{4} \\ ab &\leq \left( \frac{a + b}{2} \right) ^2 \\ \sqrt{ab} &\leq \frac{a + b}{2} \end{align*}$
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 $\displaystyle \sqrt{ab}$
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 .