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stkbarlad
Pe: 2012-09-23T10:13:55+03:00 In: In:

Inegalitati

Salut! Ma puteti ajuta va rog cu urmatoarele exercitii?

1) (x+y)/(1+x+y)<x/(1+x)+y/(1+y) x,y sunt numere reale pozitive;
2)(x+1)(y+1)(x+z)(y+z)> sau = cu 16xyz x,y,z sunt numere reale pozitive
3)(x+y+z)^2< sau = cu3(x^2+y^2+z^2)
4)x^4+y^4+z^4> sau = cu xyz(x+y+z)

Va multumesc!

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4 raspunsuri

  1. GreatMath
    veteran (III)

    2)(x+1)(y+1)(x+z)(y+z)> sau = cu 16xyz x,y,z sunt numere reale


    \begin{array}{l} \underbrace {\left( {x + 1} \right)\left( {y + 1} \right)\left( {x + z} \right)\left( {y + z} \right)}_P \ge 16xyz\\ {\left( {{\rm{MA - MG}}} \right)_2}:\\ \left\{ \begin{array}{l} x + 1 \ge 2\sqrt x \\ y + 1 \ge 2\sqrt y \\ x + z \ge 2\sqrt {xz} \\ y + z \ge 2\sqrt {yz} \end{array} \right. \Rightarrow P \ge {2^4}\sqrt {x \times y \times xz \times yz} = 16xyz \end{array}

  2. stkbarlad

    Multumesc mult!

  4. GreatMath
    veteran (III)

    3)(x+y+z)^2< sau = cu3(x^2+y^2+z^2)


    \begin{array}{l} 3\left( {{x^2} + {y^2} + {z^2}} \right) \ge {\left( {x + y + z} \right)^2}\\ {\rm{Aplicam CB}}{{\rm{S}}_3}:\\ \left( {{1^2} + {1^2} + {1^2}} \right)\left( {{x^2} + {y^2} + {z^2}} \right) \ge {\left( {1 \times x + 1 \times y + 1 \times z} \right)^2} = {\left( {x + y + z} \right)^2} \end{array}

  5. GreatMath
    veteran (III)

    4)x^4+y^4+z^4> sau = cu xyz(x+y+z)


    \begin{array}{l} \underbrace {{x^4} + {y^4} + {z^4}}_S \ge xyz\left( {x + y + z} \right)\\ {\left( {{\rm{MA - MG}}} \right)_2}:\\ {x^4} + {y^4} \ge 2{\left( {xy} \right)^2}\,\mathop \leftrightarrow \limits^{{\rm{Analog}}} \,S \ge \underbrace {{{\left( {xy} \right)}^2} + {{\left( {yz} \right)}^2} + {{\left( {zx} \right)}^2}}_{S'}\\ {\rm{Din nou }}{\left( {{\rm{MA - MG}}} \right)_2}:\\ {\left( {xy} \right)^2} + {\left( {yz} \right)^2} \ge 2x{y^2}z\,\mathop \leftrightarrow \limits^{{\rm{Analog}}} \,\,S' \ge \left( {xy} \right)\left( {yz} \right) + \left( {yz} \right)\left( {zx} \right) + \left( {zx} \right)\left( {xy} \right) = xyz\left( {x + y + z} \right) \end{array}
    Dupa cum vezi…. inegalitatea este foarte slaba 😀

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