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Thread: Is proof valid?

  1. #1
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    Is proof valid?

    I need to prove that inequality is true:

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

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

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

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

    Transforming it we get:
    [
    $\displaystyle \frac{{b - a}}{{\sqrt b }} \le \frac{{b - a}}{{\sqrt a }}$
    $\displaystyle \frac{{\sqrt a }}{{\sqrt b }} \le \frac{{b - a}}{{b - a}}$
    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:

    $\displaystyle \sqrt a + \sqrt b \le \sqrt {\frac{{a^2 }}{b}} + \sqrt {\frac{{b^2 }}{a}}
    $

    is an identity for $\displaystyle a,b>0$.

    You assume it true and deduce that:

    $\displaystyle a \le b$.

    Hence if $\displaystyle b < a$ you have a contradiction.

    RonL
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