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andrei_g
Pe: 2012-04-20T17:18:42+03:00 In: In:

Inegalitate l

Fie a_1, a_2,…,a_n apartin intervalului (1,infinit) astfel incat (a_1)*(a_2)*…*(a_n)=10^(2n). Sa se arate ca :

(lg a_1)^n +(lg a_2)^n +…+ (lg a_n)^n>=n(n+1) ,unde n apartine lui N*.

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2 raspunsuri

  1. TheodorMunteanu

    (lg a_1)^n=(1+lg a_1-1)^n\geq 1+n(lg a_1-1)
    (lg a_2)^n=(1+lg a_2-1)^n\geq 1+n(lg a_2-1)
    …
    (lg a_n)^n=(1+lg a_n-1)^n \geq 1+n(lg a_n-1)
    am folosit de n ori inegalitatea lui Bernoulli (1+a)^n\geq 1+na,a>-1
    Daca adunam inegalitatile obtinem ca suma noastra sa zicemS ,S\geq n+n(lg \frac{a_1}{10}+lg \frac{a_2}{10}+...+lg \frac{a_n}{10})=n+n^2 deoarece lg \frac{a_1a_2...a_n}{10^n}=lg 10^n=n

  2. TheodorMunteanu

    O alta metoda ar fi sa demonstrezi ca lg a_1\cdot lg a_2\cdot ...\cdot lg a_n\geq 1+n
    asta e foarte usor de demonstrat din faptul ca (1+a_1)(1+a_2)...(1+a_n)\geq 1+\sum_{k=1}^n a_k
    deci (1+lg a_1-1)...(1+lg a_2-1)\geq 1+\sum_{k=1}^n lg\frac{a_k}{10}=1+lg 10^n=1+n

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