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Math Help - Is proof valid?

  1. #1
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    Is proof valid?

    I need to prove that inequality is true:

    \sqrt a  + \sqrt b  \le \sqrt {\frac{{a^2 }}{b}}  + \sqrt {\frac{{b^2 }}{a}} for a,b>0

    Transforming it we get:
    \sqrt a  + \sqrt b  \le \frac{a}{{\sqrt b }} + \frac{b}{{\sqrt a }}
    \frac{{b - a}}{{\sqrt b }} \le \frac{{b - a}}{{\sqrt a }}
    \frac{{\sqrt a }}{{\sqrt b }} \le \frac{{b - a}}{{b - a}}
    \frac{{\sqrt a }}{{\sqrt b }} \le 1 \Leftrightarrow \sqrt a  \le \sqrt b  \Leftrightarrow a \le b

    Now, from \frac{{b - a}}{{\sqrt b }} \le \frac{{b - a}}{{\sqrt a }} we get \sqrt a (b - a) - \sqrt b (b - a) \le 0 \Leftrightarrow (b - a)(\sqrt a  - \sqrt b ) \le 0 which is true because from first transformation we get that a \le b and \sqrt a  \le \sqrt b .

    Is this proof valid?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DenMac21
    I need to prove that inequality is true:

    \sqrt a  + \sqrt b  \le \sqrt {\frac{{a^2 }}{b}}  + \sqrt {\frac{{b^2 }}{a}} for a,b>0

    Transforming it we get:
    [
    \frac{{b - a}}{{\sqrt b }} \le \frac{{b - a}}{{\sqrt a }}
    \frac{{\sqrt a }}{{\sqrt b }} \le \frac{{b - a}}{{b - a}}
    This statement is only true if b - a > 0. An example of how this fails: Set a=4 and b=1 in the original expression. This expression is correct. Your final result, though, predicts that 4 < 1. The reason is in the line above. If b - a is negative, the inequality switches direction.

    -Dan
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  3. #3
    Grand Panjandrum
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    Something is wrong.

    You wish to prove that:

    \sqrt a  + \sqrt b  \le \sqrt {\frac{{a^2 }}{b}}  + \sqrt {\frac{{b^2 }}{a}}<br />

    is an identity for a,b>0.

    You assume it true and deduce that:

     a \le b.

    Hence if b < a you have a contradiction.

    RonL
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