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Math Help - Inequality

  1. #1
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    Inequality

    hello,
    i have to proof a^p+b^p \le(a^2+b^2)^{\frac{p}{2}}for not negative real numbers and 2 \le p<\infty. I dont know how can i do that. can you help me? i really want to learn it
    redrose5
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  2. #2
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    Re: Inequality

    Use the Binomial Theorem on the right hand side
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  3. #3
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    Re: Inequality

    ok, thank you=).
    (a^p+b^p)^2\le ((a^2+b^2)^{\frac{p}{2}})^2 \rightsquigarrow(a^p+b^p)^2=a^{2p}}+2a^pb^p+b^{2p} \le ((a^2+b^2)^{\frac{p}{2}})^2=(a^2+b^2)^p=\sum_{k=0}  ^{p}\binom{p}{k}(a^2)^{p-k}(b^2)^{k}=a^{2p}+pa^{2p-2}b^2+\frac{p(p-1)}{2}a^{2p-4}b^4+\frac{p(p-1)(p-2)}{6}a^{2p-6}b^6+....+b^{2p}
    is it correct? but why is
    2a^pb^p\le pa^{2p-2}b^2+\frac{p(p-1)}{2}a^{2p-4}b^4+\frac{p(p-1)(p-2)}{6}a^{2p-6}b^6+....} correct?
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  4. #4
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    Re: Inequality

    The inequality is clearly true if a=b or if p=2.

    So, suppose that a>b and p>2, then we require

    a^{p}+b^{p}\leq (a^{2}+b^{2})^{p/2},

    a^{p}\left(1+(\frac{b}{a})^{p}\right)\leq \left[a^{2}\left(1+(\frac{b}{a})^{2}\right)\right]^{p/2},

    a^{p}\left(1+(\frac{b}{a})^{p}\right)\leq a^{p}\left(1+(\frac{b}{a})^{2}\right)^{p/2},

    and, if we let b/a=x,

    1+x^{p}\leq (1+x^{2})^{p/2}, \quad x<1.

    Now, if p>2 and x<1, then x^{p}<x^{2}, so 1+x^{p}<1+x^{2} and therefore

    1+x^{p} < (1+x^{2})^{p/2}, as required.
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