Nice inequality



  • Let a,b,ca,b,c are positive real number,abc1abc \geq 1.Prove that:
    a3a3+a2+a+1+b3b3+b2+b+1+c3c3+c2+c+134\frac{a^3}{a^3+a^2+a+1}+\frac{b^3}{b^3+b^2+b+1}+\frac{c^3}{c^3+c^2+c+1}\geq \frac{3}{4}



  • Penser à C.S et la méthode pqr



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  • This post is deleted!


  • On utilisera les notations p=a+b+c p=a+b+c et q=ab+ac+bcq=ab+ac+bc et r=abc r=abc
    D apres C.SC.S L.H.S(a+b+c)2p2+p2q+qr+3=p2p2+p2q+qr+3L.H.S\geq \frac{(a+b+c)^2}{p^2+p-2q +\frac{q}{r}+3}= \frac{p^2}{p^2+p-2q +\frac{q}{r}+3}
    Donc il suffit de montrer que 4p2 4p^2 \geq 3(p2+p2q+qr+3) 3 (p^2+p-2q +\frac{q}{r}+3)
    Equivalent a p2+6q3p+3qr+9 p^2+6q \geq 3p +3 \frac{q}{r} +9
    Or on a p3(abc)1/3=3p\geq 3(abc)^{1/3} =3 donc p23pp^2\geq 3p
    et 3q3.3(abc)2/3=9 3q \geq 3.3(abc)^{2/3} =9
    et 3q3qr3q\geq \frac{3q}{r} Car r1r\geq 1
    CQFDCQFD


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