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alina98
Pe: 2012-10-27T13:22:50+03:00 In: In:

inegalitati

Sa se demonstreze inegalitatile:

a^6+b^6+1≥3a^2b^2

2/b+c +2/c+a +2/a+b≥ 9/a+b+c

^=putere

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5 raspunsuri

  1. sandy_sc
    maestru (V)

    Aplici direct formula inegalitatii mediilor(Ma>,=Mg)
    In cazul de fata termenii sunt a^6 b^6, 1

  2. PhantomR
    expert (VI)

    Metoda 1 \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\geq \frac{9}{a+b+c} |\cdot \frac{a+b+c}{2} \Leftrightarrow \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\geq \frac{9}{2} \Leftrightarrow \left( \frac{a+b}{a+b}+\frac{c}{a+b}\right ) + \left( \frac{b+c}{b+c}+\frac{a}{b+c}\right) + \left( \frac{c+a}{c+a}+\frac{b}{c+a}\right ) \geq \frac{9}{2} \Leftrightarrow \left( 1+\frac{c}{a+b}\right)+\left( 1+\frac{1}{b+c}\right) +\left( 1+\frac{b}{c+a}\right )\geq \frac{9}{2} |-3 \Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}-3=\frac{9-6}{2}=\frac{3}{2}, care este inegalitatea lui NesbittMetoda 2inegalitatea lui Titu Andreescu avem \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}=\frac{2^2}{2(a+b)}+\frac{2^2}{2(b+c)}+\frac{2^2}{2(c+a)}\geq \frac{(2+2+2)^2}{2(a+b)+2(b+c)+2(c+a)} = \frac{36}{4(a+b+c)}=\frac{9}{a+b+c}, ceea ce trebuia demonstrat.

  4. gunty
    maestru (V)

    1.
    Din inegalitatea mediilor iese ca

        \[ \frac{{a^6 + b^6 + 1}}{3} \ge \sqrt[3]{{a^6 \cdot b^6 \cdot 1}};\forall a,b,c \in R \Leftrightarrow a^6 + b^6 + 1 \ge 3a^2 \cdot b^2 \]

    2.

        \[ \begin{array}{l} \frac{2}{{b + c}} + \frac{2}{{c + a}} + \frac{2}{{a + b}} \ge \frac{9}{{a + b + c}};a,b,c \in R_ + \\ \Leftrightarrow (2a + 2b + 2c)\left( {\frac{1}{{b + c}} + \frac{1}{{c + a}} + \frac{1}{{a + b}}} \right) \ge 9 \\ \end{array} \]

    Fie x=b+c, y=c+a, z=a+b

        \[ \begin{array}{l} \Rightarrow (x + y + z)\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) \ge 9 \Leftrightarrow 1 + \frac{x}{y} + \frac{x}{z} + \frac{y}{x} + 1 + \frac{y}{z} + \frac{z}{x} + \frac{z}{y} + 1 \ge 9 \\ \Leftrightarrow \frac{{x^2 }}{{xy}} + \frac{{x^2 }}{{xz}} + \frac{{y^2 }}{{xy}} + \frac{{y^2 }}{{yz}} + \frac{{z^2 }}{{xz}} + \frac{{z^2 }}{{yz}} \ge 6 \Leftrightarrow \frac{{x^2 + y^2 }}{{xy}} + \frac{{x^2 + z^2 }}{{xz}} + \frac{{y^2 + z^2 }}{{yz}} \ge 6 \\ \end{array} \]

    (*)

    Sa demonstram ca pentru orice

        \[ m,n \in R_ + \]

    avem inegalitatea

        \[ \frac{{m^2 + n^2 }}{{mn}} \ge 2 \]

        \[ \Leftrightarrow m^2 + n^2 \ge 2mn \Leftrightarrow m^2 - 2mn + n^2 \ge 0 \Leftrightarrow (m - n)^2 \ge 0 \]

    (adevarat)

    Deci

        \[ \frac{{x^2 + y^2 }}{{xy}} \ge 2;\frac{{x^2 + z^2 }}{{xz}} \ge 2;\frac{{y^2 + z^2 }}{{yz}} \ge 2 \]

    In concluzie, inegalitatea notata cu (*) este adevarata.

  5. alina98

    multumesc tuturor! 🙂

  6. PhantomR
    expert (VI)
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