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Pe: 2013-02-08T21:54:33+02:00 In: In:

inegalitate

OLM 2012 etapa locala Brasov:

Fie numerele a,b,c apartinand intervalului (0,1) si x,y,z>0 astfel incat a=(bc)^x, c=(ab)^z, b=(ac)^y. Aratati ca 1/(x+y+2) + 1/(y+z+2) + 1/(x+z+2) <=1.

Am scris x=logaritm in baza ab din c si analoagele.

Am ajuns apoi, folosind inegalitatea mediilor de cateva ori la 1/x + 1/y + 1/z <=1. Apoi am incercat sa scriu logaritmii la baza comuna, dar nu da ceva prea frumos.

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3 raspunsuri

  1. George_Gaumont

    Rezolvarea, probabil a autorului (Cezar Lupu, Bucuresti), din G.M. nr.3/2012
    http://postimage.org/image/6krrv5oyh/

  2. viperaza

    Multumesc!

  4. gunty
    maestru (V)

    Avem

        \[ \begin{array}{l} x,y,z > 0 \\ a,b,c \in \left( {0,1} \right) \\ a = \left( {bc} \right)^x \Rightarrow x = \log _{bc} a \Leftrightarrow x + 1 = \log _{bc} abc \Leftrightarrow \frac{1}{{x + 1}} = \log _{abc} bc_{\left( {*1} \right)} \\ b = \left( {ac} \right)^y \Rightarrow y = \log _{ac} b \Leftrightarrow y + 1 = \log _{ac} abc \Leftrightarrow \frac{1}{{y + 1}} = \log _{abc} ac_{\left( {*2} \right)} \\ c = \left( {ab} \right)^z \Rightarrow z = \log _{ab} c \Leftrightarrow z + 1 = \log _{ab} abc \Leftrightarrow \frac{1}{{z + 1}} = \log _{abc} ab_{\left( {*3} \right)} \\ \end{array} \]

    Sa demonstram urmatoarea inegalitate:

        \[ \frac{1}{4} \cdot \left( {\frac{1}{m} + \frac{1}{n}} \right) \ge \frac{1}{{m + n}};m,n > 0 \]

    .

        \[ \Leftrightarrow \frac{1}{4} \cdot \frac{{m + n}}{{mn}} \ge \frac{1}{{m + n}} \Leftrightarrow \frac{{4mn}}{{m + n}} \le m + n \Leftrightarrow 4mn \le \left( {m + n} \right)^2 \Leftrightarrow m^2 - 2mn + n^2 \ge 0 \Leftrightarrow \left( {m - n} \right)^2 \ge 0_{\left( {adev.} \right)} \]

    Trebuie aratat, ca:

        \[ \begin{array}{l} \frac{1}{{x + y + 2}} + \frac{1}{{y + z + 2}} + \frac{1}{{x + z + 2}} \le 1 \\ \Leftrightarrow \frac{1}{{\left( {x + 1} \right) + \left( {y + 1} \right)}} + \frac{1}{{\left( {y + 1} \right) + \left( {z + 1} \right)}} + \frac{1}{{\left( {x + 1} \right) + \left( {z + 1} \right)}} \le 1 \\ \end{array} \]

    Din inegalitatea demonstrata la inceput, primim:

        \[ \left. \begin{array}{l} \frac{1}{4} \cdot \left( {\frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {y + 1} \right)}}} \right) \ge \frac{1}{{\left( {x + 1} \right) + \left( {y + 1} \right)}} \\ \frac{1}{4} \cdot \left( {\frac{1}{{\left( {y + 1} \right)}} + \frac{1}{{\left( {z + 1} \right)}}} \right) \ge \frac{1}{{\left( {y + 1} \right) + \left( {z + 1} \right)}} \\ \frac{1}{4} \cdot \left( {\frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {z + 1} \right)}}} \right) \ge \frac{1}{{\left( {x + 1} \right) + \left( {z + 1} \right)}} \\ \end{array} \right\} \Rightarrow \frac{1}{2}\left( {\frac{1}{{x + 1}} + \frac{1}{{y + 1}} + \frac{1}{{z + 1}}} \right) \ge \frac{1}{{\left( {x + 1} \right) + \left( {y + 1} \right)}} + \frac{1}{{\left( {y + 1} \right) + \left( {z + 1} \right)}} + \frac{1}{{\left( {x + 1} \right) + \left( {z + 1} \right)}} \]

    Deci este destul de demonstrat, ca

        \[ \frac{1}{2}\left( {\frac{1}{{x + 1}} + \frac{1}{{y + 1}} + \frac{1}{{z + 1}}} \right) \le 1 \]

    (*1),(*2),(*3)=>Inegalitatea este echivalenta cu

        \[ \frac{1}{2}\left( {\log _{abc} bc + \log _{abc} ac + \log _{abc} ab} \right) \le 1 \Leftrightarrow \frac{1}{2}\log _{abc} \left( {abc} \right)^2 \le 1 \Leftrightarrow \frac{1}{{\not 2}} \cdot \not 2\log _{abc} abc \le 1 \Leftrightarrow 1 \le 1_{\left( {adev.} \right)} \]

    * Daca te intereseaza, iti propun o problema foarte asemanatoare (e din etapa locala a olimpiadei – Sighetu Marmatiei):

    a):
    Fie

        \[ \alpha ,\beta ,\gamma > 0 \]

    . Sa se demonstreze ca

        \[ \frac{{\alpha \beta \gamma }}{{\alpha \beta + \beta \gamma + \gamma \alpha }} \le \frac{{\alpha + \beta + \gamma }}{9} \]

    .

    b):
    Fie

        \[ a,b,c,d \in R \]

    , cu

        \[ a,b,c,d \in \left( {0,1} \right) \]

    sau

        \[ a,b,c,d \in \left( {1,\infty } \right) \]

    si

        \[ x = \log _{bcd} a \]

    ,

        \[ y = \log _{acd} b \]

    ,

        \[ z = \log _{abd} c \]

    ,

        \[ t = \log _{abc} d \]

    . Sa se arate ca

        \[ \frac{1}{{y + z + t + 3}} + \frac{1}{{x + z + t + 3}} + \frac{1}{{x + y + t + 3}} + \frac{1}{{x + y + z + 3}} \le 1 \]

    .

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