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Math Help - calc inequality

  1. #1
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    calc inequality

    My books says to use calculus and algebra to prove this inequality, assuming that x is a positive number:
    ln(1+x) less than or equal to x/(sqrt(1+x)).
    I have no idea what to do and would greatly appreciate any help. Thanks in advance.
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  2. #2
    Super Member PaulRS's Avatar
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    Consider: \frac{1}{x+1}\leq{\frac{x+2}{2\cdot{(x+1)\cdot{\sq  rt[]{x+1}}}}}

    Which clearly holds for x\geq{0}

    Just to see: \frac{1}{x+1}\leq{\frac{x+2}{2\cdot{(x+1)\cdot{\sq  rt[]{x+1}}}}}

    If and only if (since x\geq{0}) : 1\leq{\frac{x+2}{2\cdot{\sqrt[]{x+1}}}}
    Which is:
    1\leq{\frac{\sqrt[]{x+1}}{2}+\frac{1}{2\cdot{\sqrt[]{x+1}}}}

    Let: z=\sqrt[]{x+1}\geq{1}

    So: 1\leq{\frac{z}{2}+\frac{1}{2\cdot{z}}}

    Multiplying by z: z\leq{\frac{z^2}{2}+\frac{1}{2}} that is true since (z-1)^2>=0 and then zē+1>=2z

    Integrating the inequality: \int_0^b\frac{dx}{x+1}\leq{\int_0^b\frac{x+2}{2\cd  ot{(x+1)\cdot{\sqrt[]{x+1}}}}}dx for b\geq{0} What do you see?
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  3. #3
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    Is it really this difficult? We haven't gone over integrals, so I'm not really sure I understand the idea, but thanks for trying to help me. Is there something simpler (possibly involving differentation)?
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  4. #4
    Super Member PaulRS's Avatar
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    Both expressions are equal when x=0, but their derivatives have the following property: (\ln(x+1))'=\frac{1}{x+1}\leq{\frac{x+2}{2\cdot{(x  +1)\cdot{\sqrt[]{x+1}}}}}=(\frac{x}{\sqrt[]{1+x}})'
    (as shown in the other post)
    for x\geq{0}
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