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Pe: 2013-08-07T08:31:54+03:00 In: In:

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Fie x,y,z,b,c,d reale cu proprietatile: x>=0; x+y>=0; x+y+z>=0; b>=c>=d>1. Sa se arate ca pentru orice a>1 au log inegalitatile:

i) b^{a^{x}}\cdot c^{a^{y}}\geq bc

ii) b^{a^{x}}\cdot c^{a^{y}}\cdot d^{a^{z}}\geq bcd

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5 raspunsuri

  1. TheodorMunteanu

    o sa rezolv punctul a) mai intai ca la punctul b) e ceva mai migalos
    Lucram in cazul y<0
    observam ca a^x+a^y\geq 2\sqrt{a^x\cdot a^y}=2\sqrt{a^{x+y}}\geq 2,
    deci ideal ar fi sa demonstram ca b^{a^x}\cdot c^{a^y}\geq (bc)^{\frac{a^x+a^y}{2}}
    Aceasta e echivalent cu b^{a^x}\cdot c^{a^y}\geq b^{\frac{a^x+a^y}{2}}\cdot c^{\frac{a^x+a^y}{2}}\Leftrightarrow b^{\frac{a^x}{2}}\cdot c^{\frac{a^y}{2}}\geq b^{\frac{a^y}{2}}\cdot c^{\frac{a^x}{2}}care e echivalent cu b^{a^x}\cdot c^{a^y}\geq b^{a^y}\cdot c^{a^x}
    sau \left(\frac{b}{c}\right)^{a^x}\geq \left(\frac{b}{c}\right)^{a^y}care e adevarata deoarece y<0 si \frac{b}{c}>1

    in cazul in care y>0,b^{a^x}\geq b,c^{a^y}\geq c de unde rezulta imediat inegalitatea a)

  2. TheodorMunteanu

    observam ca \frac{a^x+a^y+a^z}{3}\geq \sqrt[3]{a^{x+y+z}}\geq 1
    si ar trebui sa demonstram deci ca b^{a^x}\cdot c^{a^y}\cdot d^{a^z}\geq (bcd)^{\frac{a^x+a^y+a^z}{3}}
    pentru aceasta logaritmam inegalitatea si avem \log b^{a^x}+\log c^{a^y}+\log d^{a^z}=a^x\log b+a^y\log c+a^z\log d\geq \frac{a^x+a^y+a^z}{3}(\log b+\log c+\log d)
    inegalitate echivalenta cu 3(a^x \log b+a^y \log c+a^z\log d)\geq (a^x+a^y+a^z)(\log b+\log c+\log d)care e adevarata din inegalitatea lui cebasev.

    PS:daca stau bine sa ma gandesc,si la inegalitatea i) mergea o logaritmare si o inegalitate cebasev(sau a rearanjarii) ca solutie alternativa.

  4. PhantomR
    expert (VI)

  5. TheodorMunteanu

    cam ai dreptate,trebuie sa gasesc o alta varianta,pe cat posibil,necalculatorie!

  6. gunty
    maestru (V)

    ii)

        \[ \begin{array}{l} b^{a^x } \cdot c^{a^y } \cdot d^{a^z } \ge bcd \\ \Leftrightarrow S = \left( {a^x - 1} \right)\lg b + \left( {a^y - 1} \right)\lg c + \left( {a^z - 1} \right)\lg d \ge 0 \\ \end{array} \]

        \[ \left. \begin{array}{l} b \ge c \ge d > 1 \Leftrightarrow \lg b \ge \lg c \ge \lg d > 0 \\ \left. \begin{array}{l} x \ge 0 \\ a > 1 \\ \end{array} \right\} \Rightarrow a^x - 1 \ge 0 \\ \end{array} \right\} \Rightarrow \left( {a^x - 1} \right)\lg b \ge \left( {a^x - 1} \right)\lg c \Rightarrow S \ge P = \left( {a^x + a^y - 2} \right)\lg c + \left( {a^z - 1} \right)\lg d \]

        \[ \left. \begin{array}{l} Ma - Mg \Rightarrow a^x + a^y \ge 2\sqrt {a^{x + y} } \ge 2\sqrt {a^0 } = 2 \Rightarrow a^x + a^y - 2 \ge 0 \\ \lg b \ge \lg c \ge \lg d > 0 \\ \end{array} \right\} \Rightarrow \left( {a^x + a^y - 2} \right)\lg c \ge \left( {a^x + a^y - 2} \right)\lg d \Rightarrow P \ge \left( {a^x + a^y + a^z - 3} \right)\lg d \]

        \[ \begin{array}{l} Deoarece\;S \ge P \ge \left( {a^x + a^y + a^z - 3} \right)\lg d,\;este\;destul\;sa\;aratam\;ca\quad \left( {a^x + a^y + a^z - 3} \right)\lg d \ge 0_{\left( 1 \right)} . \\ \left. \begin{array}{l} Ma - Mg \Rightarrow a^x + a^y + a^z \ge 3\sqrt[3]{{a^{x + y + z} }} \ge 3\sqrt[3]{{a^0 }} = 3 \Rightarrow a^x + a^y + a^z - 3 \ge 0 \\ \lg d > 0 \\ \end{array} \right\} \Rightarrow \left( 1 \right)adev. \\ \end{array} \]

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