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Alexandru10
Pe: 2022-09-20T19:15:26+03:00 In: In:

Inegalitati

a,b,c\epsilonR+
a^{2}+b^{2}+c^{2}=1
atunci -1/2\leqab+bc+ca\leq1

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4 raspunsuri

  1. red12dog34
    veteran (III)

    Avem
    0\le (a+b+c)^{2}=a^2+b^2+c^2+2ab+2ac+2bc\Rightarrow\\ \Rightarrow 2(a+b+c)\ge -1\Rightarrow ab+ac+bc\ge-\frac{1}{2}
    Avem inegalitatea a^2+b^2+c^2\ge ab+ac+bc
    Rezultă ab+ac+bc\le 1

  2. Alexandru10

    red_dog post_id=116735 time=1663744741 user_id=3357 wrote:
    \Rightarrow\\ 1\Rightarrow 2(a+b+c)> -1\[/tex]


    Multumesc foarte mult , dar n-m inteles partea de sus.De ce-i asa?

  4. red12dog34
    veteran (III)

    red_dog post_id=116735 time=1663744741 user_id=3357 wrote:
    Avem
    0\le (a+b+c)^{2}=a^2+b^2+c^2+2ab+2ac+2bc\Rightarrow\\ \Rightarrow 2(ab+bc+ca)\ge -1\Rightarrow ab+ac+bc\ge-\frac{1}{2}
    Avem inegalitatea a^2+b^2+c^2\ge ab+ac+bc
    Rezultă ab+ac+bc\le 1


    Da, am scris eu greșit. Am corectat.

  5. Alexandru10

    Multumesc foarte mult!
    🙂

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