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dennis9091
guru (IV)
Pe: 2013-01-04T11:26:33+02:00 In: In:

Inegalitati

Sa se determine numerele intregi a,b,c stiind ca satisfac relatiie :
a^2<2a+b-c
b^2<2b+c-a
c^2<2c+a-b
(RMT 2/1985).
PS: Am incercat sa le adun .. dar nu am oobtinut mai nimic.. nu am reusit sa fac suma de patrate.

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  1. gunty
    maestru (V)

    Am pornit de la ideea ta de a le aduna:

    *** QuickLaTeX cannot compile formula:
\[
	\begin{array}{l}
	 \left. {\begin{array}{   }
	   {a^2  < 2a + b - c}  \\
	   {b^2  < 2b + c - a}  \\
	   {c^2  < 2c + a - b}  \\
	\end{array}} \right\} +  \\
	  \Rightarrow a^2  + b^2  + c^2  < 2a + b - c + 2b + c - a + 2c + a - b \\
	  \Leftrightarrow a^2  + b^2  + c^2  < 2a + 2b + 2c \Leftrightarrow \left( {a^2  - 2a} \right) + \left( {b^2  - 2b} \right) + \left( {c^2  - 2c} \right) < 0 \\
	  \Leftrightarrow \left( {a^2  - 2a + 1} \right) + \left( {b^2  - 2b + 1} \right) + \left( {c^2  - 2c + 1} \right) < 3 \\
	  \Leftrightarrow \left( {a - 1} \right)^2  + \left( {b - 1} \right)^2  + \left( {c - 1} \right)^2  < 3 \\
	 \end{array}
	\]

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    Pentru ca inegalitatea primita sa fie adevarata, ai urmatoarele posibilitati:

        \[ \begin{array}{l} a = 0,b = 0,c = 1 \\ a = 0,b = 1,c = 0 \\ a = 0,b = 1,c = 1 \\ a = 0,b = 1,c = 2 \\ a = 0,b = 2,c = 1 \\ a = 1,b = 0,c = 0 \\ a = 1,b = 0,c = 1 \\ a = 1,b = 0,c = 2 \\ a = 1,b = 1,c = 0 \\ a = 1,b = 1,c = 1 \\ a = 1,b = 1,c = 2 \\ a = 1,b = 2,c = 0 \\ a = 1,b = 2,c = 1 \\ a = 1,b = 2,c = 2 \\ a = 2,b = 0,c = 1 \\ a = 2,b = 1,c = 0 \\ a = 2,b = 1,c = 1 \\ a = 2,b = 1,c = 2 \\ a = 2,b = 2,c = 1 \\ \end{array} \]

    Trebuie verificate toate posibilitatile, daca indeplinesc cele trei inegalitati date la inceput. Cele care le indeplinesc, normal ca le luam solutie, iar cele care nu, nu le luam in considerare.

    * E o rezolvare cam lunga, si nu prea „eleganta”, insa momentan nu stiu o alta metoda.

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