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AkamiKitsune
Pe: 2012-10-22T14:37:29+03:00 In: In:

Exercitii algebra (urgent)

Hey, stie care-va sa-mi rezolve si mie exercitiile astea? Le am ca tema la matematica dar nu inteleg cum trebuie sa le fac…

Demonstrati inegalitatile:
1. a+1/a mai mare sau egal cu 2

2. (b+c)/a + (a+c)/b + (a+B)/c mai mare sau egal cu 6

3. a(b^2+c^2) + b(a^2+c^2) + c(a^2+b^2) mai mare sau egal cu 6abc

Sunt mai multe exercitii vreo 15 dar sper ca dupa ce voi vedea rezolvarea la cateva le voi putea face singura…

Multumesc mult!

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2 raspunsuri

  1. dennis9091
    guru (IV)

    La prima inmultesti totu cu a si le dai in stanga si faci patrat perfect care e mai mare sau egal cu 0. La A doua Folosesti prima inegalitate..

  2. gunty
    maestru (V)

    1.

        \[ \begin{array}{l} a + \frac{1}{a} \ge 2;a \in R_ + \\ \Leftrightarrow \frac{{a^2 + 1}}{a} \ge 2 \Leftrightarrow a^2 + 1 \ge 2a \Leftrightarrow a^2 - 2a + 1 \ge 0 \Leftrightarrow (a - 1)^2 \ge 0 \\ \end{array} \]

    2.

        \[ \begin{array}{l} \frac{{b + c}}{a} + \frac{{a + c}}{b} + \frac{{a + b}}{c} \ge 6;a,b,c \in R_ + \\ \Leftrightarrow \frac{b}{a} + \frac{c}{a} + \frac{a}{b} + \frac{c}{b} + \frac{a}{c} + \frac{b}{c} \ge 6 \\ \Leftrightarrow \frac{{b^2 }}{{ab}} + \frac{{c^2 }}{{ac}} + \frac{{a^2 }}{{ab}} + \frac{{c^2 }}{{bc}} + \frac{{a^2 }}{{ac}} + \frac{{b^2 }}{{bc}} \ge 6 \\ \Leftrightarrow \frac{{a^2 + b^2 }}{{ab}} + \frac{{a^2 + c^2 }}{{ac}} + \frac{{b^2 + c^2 }}{{bc}} \ge 6 \\ \end{array} \]

    Fie

        \[ x,y \in R_ + \]

    Sa demonstram ca

        \[ \frac{{x^2 + y^2 }}{{xy}} \ge 2 \]

        \[ \begin{array}{l} \frac{{x^2 + y^2 }}{{xy}} \ge 2 \Leftrightarrow x^2 + y^2 \ge 2xy \Leftrightarrow x^2 - 2xy + y^2 \ge 0 \Leftrightarrow (x - y)^2 \ge 0 \\ Deci \\ \frac{{a^2 + b^2 }}{{ab}} \ge 2;\frac{{a^2 + c^2 }}{{ac}} \ge 2;\frac{{b^2 + c^2 }}{{bc}} \ge 2 \\ \end{array} \]

    3.

        \[ \begin{array}{l} a(b^2 + c^2 ) + b(a^2 + c^2 ) + c(a^2 + b^2 ) \ge 6abc \\ \Leftrightarrow ab^2 + ac^2 + ba^2 + bc^2 + ca^2 + cb^2 \ge 6abc \\ \end{array} \]

    Fie

        \[ x,y,z \in R_ + \]

    Sa demonstram ca

        \[ xy^2 + xz^2 \ge 2xyz \]

    Stim ca

        \[ \frac{{xy^2 + xz^2 }}{2} \ge \sqrt {xy^2 \cdot xz^2 } \]

        \[ \begin{array}{l} \Leftrightarrow xy^2 + xz^2 \ge 2\sqrt {x^2 y^2 z^2 } \Leftrightarrow xy^2 + xz^2 \ge 2xyz \\ Deci \\ ab^2 + ac^2 \ge 2abc \\ ba^2 + bc^2 \ge 2abc \\ ca^2 + cb^2 \ge 2abc \\ \end{array} \]

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