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Nguyễn Hưng Phát

27/11/2017 at 14:37
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Givea>b>c>0 and a2+b2+c2=1.Prove that:\(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\ge\dfrac{1}{2}\)




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  • ...
    Alchemy 27/11/2017 at 19:32

    By Cauchy-Schwarz's inequality:

    \(L.H.S=\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\)

    \(=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)

    \(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}\)

    \(\ge\dfrac{a^2+b^2+c^2}{2}=\dfrac{1}{2}=R.H.S\)

    DONE!

    Nguyễn Hưng Phát selected this answer.
  • ...
    Nguyễn Hiền 29/11/2017 at 08:40

    L . H . S = a 3 b + c+ b 3 c + a+ c 3 a + bL.H.S=a3b+c+b3c+a+c3a+b

    = a 4 a b + a c+ b 4 b c + a b+ c 4 a c + b c=a4ab+ac+b4bc+ab+c4ac+bc

    ≥ ( a 2 + b 2 + c 2 ) 2 2 ( a b + b c + c a )≥ ( a 2 + b 2 + c 2 ) ( a b + b c + c a ) 2 ( a b + b c + c a )≥(a2+b2+c2)22(ab+bc+ca)≥(a2+b2+c2)(ab+bc+ca)2(ab+bc+ca)

    ≥ a 2 + b 2 + c 2 2= 1 2= R . H . S

  • ...
    KEITA FC 8C 28/11/2017 at 21:34

    By Cauchy-Schwarz's inequality:

    L.H.S=a3b+c+b3c+a+c3a+bL.H.S=a3b+c+b3c+a+c3a+b

    =a4ab+ac+b4bc+ab+c4ac+bc=a4ab+ac+b4bc+ab+c4ac+bc

    ≥(a2+b2+c2)22(ab+bc+ca)≥(a2+b2+c2)(ab+bc+ca)2(ab+bc+ca)≥(a2+b2+c2)22(ab+bc+ca)≥(a2+b2+c2)(ab+bc+ca)2(ab+bc+ca)

    ≥a2+b2+c22=12=R.H.S


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