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dana1101
Pe: 2013-05-09T16:17:19+03:00 In: In:

Inegalitate

Demonstrati inegalitatea pentru oricare n natural mai mare sau egal cu 3:
]√(de ordin n din)(n)]+(5)/(3n)<2

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  1. ghioknt
    profesor

    Daca am inteles bine, trebuie demonstrat ca

        \[ \sqrt[n]{n} + \frac{5}{{3n}} < 2\,\,sau\,\,\sqrt[n]{n} < 2 - \frac{5}{{3n}} \]

    .

        \[ \sqrt[n]{n} = \sqrt[n]{{\frac{{n!}}{{\left( {n - 1} \right)!}}}} = \sqrt[n]{{1 \cdot \frac{2}{1} \cdot \frac{3}{2} \cdot ... \cdot \frac{n}{{n - 1}}}} < \frac{{1 + \frac{2}{1} + \frac{3}{2} + ... + \frac{n}{{n - 1}}}}{n} \]

    (ineg. stricta intre cele 2 medii). Demonstratia e terminata daca aratam ca 

    *** QuickLaTeX cannot compile formula:
\[
	\begin{array}{l}
	 \frac{{1 + \frac{2}{1} + \frac{3}{2} + ... + \frac{n}{{n - 1}}}}{n} < 2 - \frac{5}{{3n}}\,\,sau\,\,1 + \frac{2}{1} + \frac{3}{2} + ... + \frac{n}{{n - 1}} < 2n - \frac{5}{3}\,\,\,pe\,\,care\,\,o\,\,notam\,\,P\left( n \right) \\
	 P\left( 4 \right)]
	

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    adevarat.
    Fie P(n) una dintre propozitiile adevarate, din care, adunand (n+1)/n in ambii membri, obtinem inegalitatea adevarata

        \[ \begin{array}{l} \,1 + \frac{2}{1} + \frac{3}{2} + ... + \frac{n}{{n - 1}} + \frac{{n + 1}}{n} < 2n - \frac{5}{3} + \frac{{n + 1}}{n}\,\,\,\,(1) \\ 2n - \frac{5}{3} + \frac{{n + 1}}{n}\, < 2\left( {n + 1} \right) - \frac{5}{3}\,\,\,(2)\,\,\,\,\, \Leftrightarrow \frac{{n + 1}}{n}\, < 2 \\ \end{array} \]

    , deci si (2) este adev. Din (1) si (2) deducem:
    P(n+1) este adev. daca P(n) este adev. oricare ar fi n>=4; cum P(4) este adev. am demonstrat ca toate inegalit. P(n) sunt adev. pt. n>=4.
    Pentru n=3 se verifica direct ine galitatea din enunt.
    PS Dupa ce am terminat aceasta demonstratie elaborata, imi dau seama de simplitatea problemei: se cam stie ca sirul n^(1/n)
    este descrescator de la 3 in sus si la fel si 5/(3n); deci daca inegalitatea se verifica pentru n=3, cu atat mai mult ea este adev. pt.n>3!!!
    Cu bine, ghioknt.

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