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Math Help - Help with an algebraic proof

  1. #1
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    Help with an algebraic proof

    Prove that if 0 \leq a<b, then

    a < \sqrt{ab} < \frac{a+b}{2} < b

    I know how to find a < \sqrt{ab}, but I don't understand how to find \sqrt{ab} < \frac{a+b}{2}, or that that will be less than b.

    Can this even be proved? It doesn't appear to be true.
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\begin{gathered}<br />
  ab < \left( {\frac{{a + b}}<br />
{2}} \right)^2  \hfill \\<br />
   \Leftrightarrow 4ab < a^2  + 2ab + b^2  \hfill \\<br />
   \Leftrightarrow 0 < \left( {a - b} \right)^2  \hfill \\ <br />
\end{gathered}
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  3. #3
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    The last part first.
    a < b \Rightarrow \quad \frac{a}{2} < \frac{b}{2} \Rightarrow \quad \frac{a}{2} + \frac{a}{2} < \frac{b}{2} + \frac{a}{2}\;\& \quad \frac{a}{2} + \frac{b}{2} < \frac{b}{2} + \frac{b}{2}.

    Now the first:
    Lemma: \left( {\forall x\,\& \,y} \right)\left[ {x^2  + y^2  \ge 2xy} \right].
    Proof: \left( {x - y} \right)^2  \ge 0 \Rightarrow \quad x^2  - 2xy + y^2  \ge 0 \Rightarrow \quad x^2  + y^2  \ge 2xy

    Now let x = \sqrt a \;\& \; y = \sqrt b and divide by 2.
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  4. #4
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    Here's a geometric proof.

    If we try to construct a square equal in area to a given rectangle and the

    dimensions of the rectangle are a and b, then the problem is to determine x

    such that x^{2}=ab.

    Look at the semi-circle in the diagram. The inscribed triangle is a right triangle, then the two smaller ones are similar. Then, we have:

    \frac{a}{x}=\frac{x}{b}

    Therefore, x is mean proportional between a and b, so there geometric mean is \sqrt{ab}.

    Since the radius of the semi-circle is \frac{a+b}{2}, it follows that:

    \sqrt{ab}\leq\frac{a+b}{2}.

    This holds even when a=b.

    The geometric mean of two positive numbers never exceeds their arithmetic mean

    Another thing to look at algebraically is since

    (a+b)^{2}-(a-b)^{2}=4ab, then

    \frac{a+b}{2}\geq\sqrt{ab}
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  5. #5
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    Thanks for the proofs - they were very helpful.

    Quote Originally Posted by Plato View Post
    Now the first:
    Lemma: \left( {\forall x\,\& \,y} \right)\left[ {x^2  + y^2  \ge 2xy} \right].
    Hmm, I don't recognize that notation.

    Does anyone know of a good book that explains/gives practice with proofs? I don't need to for any of my courses (I'm currently in Calc II), but I think proofs are interesting, and I'd like to get better at them. Sadly I have little experience because none of the math courses I've taken required me to do proofs. Even when I took geometry in high school, they dumbed it down and barely required any proofs. It's kind of unfortunate, because I'd have more mathematical discipline if it was for a grade. Anyway, I would be interested to learn about proofs in my spare time.
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  6. #6
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    Quote Originally Posted by paulrb View Post
    Hmm, I don't recognize that notation.
    \left( {\forall x\,\& \,y} \right)
    That is read "For all x & y" what follows is true.
    The up-side-down A is read For all.
    The backwards E, \exists is read "There exist" or "for some".
    Both notations are common in mathematical proofs.
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