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Math Help - Inequality proof.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Inequality proof.

    Prove that (x+1)\ln(x+1)+e^y\geq (1+x)(1+y), where x,y \in \mathbb{R}^+
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  2. #2
    Super Member girdav's Avatar
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    Re: Inequality proof.

    We fix x\geq 0. Let f(y) := (x+1)\ln(x+1)+e^y-(1+x)(1+y). We have f'(y) =e^y-(1+x) hence f is decreasing on \left[0,\ln(1+x)\right] and increasing on \left[\ln(1+x),+\infty[. We get that f gets its minimum at y =\ln (1+x).
    Since \begin{align*}f(\ln(1+x)) &= (x+1)\ln(1+x)+1+x-(1+x)(1+\ln(1+x)) \\&= (1+x)(\ln(1+x)+1-(1+\ln(1+x)))\\& =0\end{align*}
    the result is shown.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Inequality proof.

    Quote Originally Posted by girdav View Post
    We fix x\geq 0. Let f(y) := (x+1)\ln(x+1)+e^y-(1+x)(1+y). We have f'(y) =e^y-(1+x) hence f is decreasing on \left[0,\ln(1+x)\right] and increasing on \left[\ln(1+x),+\infty[. We get that f gets its minimum at y =\ln (1+x).
    Since \begin{align*}f(\ln(1+x)) &= (x+1)\ln(1+x)+1+x-(1+x)(1+\ln(1+x)) \\&= (1+x)(\ln(1+x)+1-(1+\ln(1+x)))\\& =0\end{align*}
    the result is shown.



    "My way":
    Spoiler:
    Take the function f(t)=\ln(t+1) and use Young's inequality - Wikipedia, the free encyclopedia
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