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

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

how would I go about proving this?

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

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

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

Originally Posted by kingsolomonsgrave
how would I go about proving this?
$sqrt(ab) \geq (a+b)/2$
i manipulated the expression so that $0 \geq (a-b)^2$ but im not sure if that proves anything
You have it written incorrectly.

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

If $a\ge 0~\&~b\ge 0$ then let $x=\sqrt a~\&~b=\sqrt b$.

3. ## Re: prove srt(ab) => (a+b)/2

Originally Posted by kingsolomonsgrave
how would I go about proving this?

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

i manipulated the expression so that $0 \geq (a-b)^2$ but im not sure if that proves anything
Are you sure your original equation wasn't \displaystyle \begin{align*} \sqrt{ab} \leq \frac{a + b}{2} \end{align*}?

4. ## Re: prove srt(ab) => (a+b)/2

yes! I knew something was off. It should have been:

it should have been:

Prove $\sqrt{ab}\leq \frac{a+b}{2}$ for all positive integers.

5. ## Re: prove srt(ab) => (a+b)/2

Originally Posted by kingsolomonsgrave
yes! I knew something was off. It should have been:

it should have been:

Prove $\sqrt{ab}\leq \frac{a+b}{2}$ for all positive integers.
\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*}

6. ## Re: prove srt(ab) => (a+b)/2

geometrically

7. ## Re: prove srt(ab) => (a+b)/2

How do we know the vertical line is sqrt(ab)?

8. ## Re: prove srt(ab) => (a+b)/2

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.

9. ## Re: prove srt(ab) => (a+b)/2

Thanks! How does that tell me that that length is $\sqrt{ab}$?

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 $a^2+a^2= \sqrt{2}$ and $b^2+b^2 = \sqrt{2}$ and $a=b so a^2=ab=b^2$ so the line is $\sqrt{ab}$

10. ## Re: prove srt(ab) => (a+b)/2

Originally Posted by kingsolomonsgrave
How does that tell me that that length is $\sqrt{ab}$?
If the altitude of the big triangle is h, then it follows from the similarity of small triangles that a / h = h / b.

11. ## Re: prove srt(ab) => (a+b)/2

Originally Posted by kingsolomonsgrave
Thanks! How does that tell me that that length is $\sqrt{ab}$?
Have you heard the term mean proportional ?
If $\frac{a}{h}=\frac{h}{b}>0$ then $h$ is the mean proportional between $a~\&~b$.

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 .