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Math Help - how to prove the inequality (1-x^2)^n>=1-nx^2?

  1. #1
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    how to prove the inequality (1-x^2)^n>=1-nx^2?

    Let x\in [0,1] and n be any positive integer, prove (1-x^2)^n\geq1-nx^2.
    It can be easily proved by induction on n, but I would like to use binomial theorem to prove it. By binomial theorem, (1-x^2)^n=1-nx^2+.... The proof is completed if the remaining "..." part in the above expansion can be proved to be \geq0, but I have no idea how to prove it. Could anyone help me with this problem? Thanks!
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  2. #2
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    Hello,

    As you said, we have :

    (1-x^2)^n=\sum_{k=0}^n {n\choose k} (-x^2)^{k}=1-nx^2+\sum_{k=2}^n {n\choose k} (-x^2)^{k}

    Now let's consider S=\sum_{k=2}^n {n\choose k} (-x^2)^{k}
    Try to make the sum start by 0, but not by adding terms, otherwise we'd be back to the beginning.
    Instead, let j=k-2

    Thus S=\sum_{j=0}^{n-2} {n\choose j+2} (-x^2)^{j+2}

    S=\sum_{j=0}^{n-2} \frac{n!}{(j+2)!(n-j-2)!} \cdot (-x^2)^{j+2}

    Now factor out some things in order to get a binomial expansion :

    S=n(n-1)(-x^2)^2\sum_{j=0}^{n-2} \frac{(n-2)!}{j!(n-2-j)!}\cdot (-x^2)^j

    S=n(n-1)x^4\sum_{j=0}^{n-2} {n-2 \choose j} (-x^2)^j

    Which is exactly n(n-1)x^4(1-x^2)^{n-2}

    And this is positive, since 1-x^2\geq 0


    Looks clear to you ?
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  3. #3
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    Thank you but, during the factoring, how to turn the (j+2)! in the denominator into j! ?
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