[niv0] inégalités de base


  • Math&Maroc

    montrer les inégalités suivantes: a,b,c0a,b,c \geq0
    a2+b2+c2ab+ac+bca^2+b^2+c^2 \geq ab+ac+bc
    3(a2+b2+c2)(a+b+c)23(a^2+b^2+c^2)\geq (a+b+c)^2
    a3+b3+c33abca^3+b^3+c^3\geq 3abc
    1a+1b4a+b\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}
    (a+b+c)(1a+1b+1c)9(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9



    1. On sait que a2+b22aba^2+b^2\geq2ab
      De même pour b et c (symétrique)
      Alors a2+b2+a2+c2+b2+c22ab+2bc+2aca^2+b^2+a^2+c^2+b^2+c^2\geq2ab+2bc+2ac
      Donc 2(a2+b2+c2)2(ab+bc+ac)2(a^2+b^2+c^2)\geq2(ab+bc+ac)
      D’où a2+b2+c2ab+bc+aca^2+b^2+c^2\geq ab+bc+ac
    2. On a (a2+b2+c2)(ab+bc+ac)(a^2+b^2+c^2)\geq(ab+bc+ac)
      Donc 2(a2+b2+c2)2(ab+bc+ac)2(a^2+b^2+c^2)\geq2(ab+bc+ac)
      Alors 2(a2+b2+c2)+(a2+b2+c2)2(ab+bc+ac)+(a2+b2+c2)2(a^2+b^2+c^2)+(a^2+b^2+c^2)\geq2(ab+bc+ac)+(a^2+b^2+c^2)
      D’où 3(a2+b2+c2)(a+b+c)23(a^2+b^2+c^2)\geq(a+b+c)^2
    3. On a (a3+b3+c3)3abc=(a+b+c)(a2+b2+c2(ab+ac+bc))(a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-(ab+ac+bc))
      Donc (a3+b3+c3)3abc0(a^3+b^3+c^3)-3abc\geq0 car a;b;c0a;b;c\geq0 et a2+b2+c2ab+bc+aca^2+b^2+c^2 \geq ab+bc+ac
      Alors (a3+b3+c3)3abc(a^3+b^3+c^3)\geq3abc
    4. On a a+bab4a+b=(a+b)24abab(a+b)\dfrac{a+b}{ab}-\dfrac{4}{a+b}=\dfrac{(a+b)^2-4ab}{ab(a+b)}
      Donc a+bab4a+b=(ab)2ab(a+b)0\dfrac{a+b}{ab}-\dfrac{4}{a+b}=\dfrac{(a-b)^2}{ab(a+b)}\geq0
      Alors 1a+1b4a+b\dfrac{1}{a}+\dfrac{1}{b}\geq\dfrac{4}{a+b}
    5. On a (a+b+c)(1a+1b+1c)=1+ab+ac+ba+1+bc+ca+cb+1(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1
      Donc, d'après l'inégalité a+1a2a+\dfrac{1}{a}\geq2, on aura ab+ba+ac+ca+bc+cb6\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}\geq6
      c.a.d ab+ba+ac+ca+bc+cb+39\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+3\geq9
      Alors (a+b+c)(1a+1b+1c)9(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\geq9

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