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mangul
Pe: 2011-03-14T20:27:02+02:00 In: In:

algebra cls 7

aratati:

    \[ \begin{array}{l} 1)x^2 + y^2 + z^2 \ge xy + xz + yz,pentru - orice - x,y,z \in R \\ 2){\textstyle{{x + y} \over z}} + {\textstyle{{y + z} \over x}} + {\textstyle{{z + x} \over y}} \ge 6;pentru - orice - x,y,z > 0 \\ 3){\textstyle{{xy} \over z}} + {\textstyle{{yz} \over x}} + {\textstyle{{zx} \over y}} \ge x + y + z;pentru - orice - x,y,z > 0 \\ \end{array} \]

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2 raspunsuri

  1. Alin

    1) x^2+y^2+z^2 \ge xy+xz+yz \Longleftrightarrow 2x^2+2y^2+2z^2\ge 2xy+2xz+2yz \Longleftrightarrow (x-y)^2+(y-z)^2+(x-z)^2\ge 0
    2) \frac{x+y}{z}+ \frac{y+z}{x}+ \frac{z+x}{y} \ge 6 \Longleftrightarrow \frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{y}+\frac{y}{x} \ge 6 .
    Stim ca \forall a,b \in (0,\infty) \Longrightarrow \frac{a}{b}+\frac{b}{a} \ge 2 (AM-GM) si aplicand de 3 ori in inegalitatea de mai sus avem concluzia.
    3)\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y} \ge x+y+z \Longleftrightarrow (xy)^2+(yz)^2+(zx)^2 \ge (xy)(xz)+(xy)(yz)+(yz)(xz) si notand xy=a;\ yz=b;\ xz=c inegalitatea devine 1).

  2. mangul

    MENTIONEZ CA SUNT PARINTE.
    AM REUSIT SA LE FAC SI EU.
    PE 2) SI 3), DUPA CE AM ADUS LA ACELASI NUMITOR, MI-A IESIT TOT CA LA 1), O SUMA DE 3 PATRATE >=0
    MULTUMESC!

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