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Math Help - prove srt(ab) => (a+b)/2

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

    Quote Originally Posted by kingsolomonsgrave View Post
    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.
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  3. #3
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    Re: prove srt(ab) => (a+b)/2

    Quote Originally Posted by kingsolomonsgrave View Post
    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*}?
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    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.
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    Re: prove srt(ab) => (a+b)/2

    Quote Originally Posted by kingsolomonsgrave View Post
    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*}
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    Re: prove srt(ab) => (a+b)/2

    geometrically
    Attached Thumbnails Attached Thumbnails prove srt(ab) => (a+b)/2-22222222222.jpg  
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    Re: prove srt(ab) => (a+b)/2

    How do we know the vertical line is sqrt(ab)?
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    Re: prove srt(ab) => (a+b)/2

    Quote Originally Posted by kingsolomonsgrave View Post
    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.
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    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}
    Last edited by kingsolomonsgrave; May 3rd 2013 at 01:19 PM.
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    Re: prove srt(ab) => (a+b)/2

    Quote Originally Posted by kingsolomonsgrave View Post
    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.
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    Re: prove srt(ab) => (a+b)/2

    Quote Originally Posted by kingsolomonsgrave View Post
    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 .
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