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Math Help - logarithms

  1. #1
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    logarithms

    Given two real numbers a and b where a>1 and b>1, prove that
    <br />
\frac{1}{\log_b a}+\frac{1}{\log_a b}\geq 2

    Attempt:

    (\log_a b)(\log_b a)=1

    <br />
\frac{1}{\log_b a}+\frac{1}{\log_a b}

    =\frac{\log_a b+\log_b a}{(\log_a b)(\log_b a)}

    I have to prove that the numerator is >=2 but i just don see a way..
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  2. #2
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    Quote Originally Posted by hooke View Post
    Given two real numbers a and b where a>1 and b>1, prove that
    <br />
\frac{1}{\log_b a}+\frac{1}{\log_a b}\geq 2

    Attempt:

    (\log_a b)(\log_b a)=1

    ...
    1. Re-write the given inequality:

    <br />
\frac{1}{\log_b( a)}+\log_b(a)\geq 2

    2. Since a > 1 and b > 1 the logarithms are greater than zero, that means they are positive. Multiply the inequality by \log_b(a):

    1+\left(\log_b(a)\right)^2\geq 2 \cdot \log_b(a)

    3. You'll get:

    \left(\log_b(a)\right)^2- 2 \cdot \log_b(a) +1 \geq 0

    \left(\log_b(a)-1\right)^2\geq 0

    4. A square is allways greater or equal zero.
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  3. #3
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    Quote Originally Posted by earboth View Post
    1. Re-write the given inequality:

    <br />
\frac{1}{\log_b( a)}+\log_b(a)\geq 2

    2. Since a > 1 and b > 1 the logarithms are greater than zero, that means they are positive. Multiply the inequality by \log_b(a):

    1+\left(\log_b(a)\right)^2\geq 2 \cdot \log_b(a)

    3. You'll get:

    \left(\log_b(a)\right)^2- 2 \cdot \log_b(a) +1 \geq 0

    \left(\log_b(a)-1\right)^2\geq 0

    4. A square is allways greater or equal zero.
    Thanks earboth, i thought of this too but after pondering a while, shouldn't we prove from any side until we get the result instead of proving that the result is true.

    Take trigonometric identities for example,

    I am supposed to prove from either side so that it equals the other side.
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