Inequality

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This topic contains 2 replies, has 3 voices, and was last updated by  Sonnhard Graubner 6 months ago.

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  • #24712

    swastik pramanik
    Participant

    \(a, b, c>0\) and \(a+b+c=1\). Prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3(a^2+b^2+c^2)$$

    #24736

    Shahbaz Khan
    Participant

    Use Cauchy-Schwarz on a/b b/c c/a and a,b,c. Should follow easily from this.

    #25147

    Sonnhard Graubner
    Participant

    Multiplying by $$abc>0$$ we have to prove that

     

    $$a^3c+ab^3+bc^3\geq 3abc(a^2+b^2+c^2)$$. Since we have

    $$a+b+c=1$$ we get

    $$(a^3c+ab^3+bc^3)(a+b+c)\geq 3abc(a^2+b^2+c^2)$$

    Using the substitution $$b=a+u,c=a+u+v$$ we get

    $$4\,{a}^{3}{u}^{2}+4\,{a}^{3}uv+4\,{a}^{3}{v}^{2}+9\,{a}^{2}{u}^{3}+18
    \,{a}^{2}{u}^{2}v+15\,{a}^{2}u{v}^{2}+3\,{a}^{2}{v}^{3}+7\,a{u}^{4}+20
    \,a{u}^{3}v+21\,a{u}^{2}{v}^{2}+8\,au{v}^{3}+a{v}^{4}+2\,{u}^{5}+7\,{u
    }^{4}v+9\,{u}^{3}{v}^{2}+5\,{u}^{2}{v}^{3}+u{v}^{4}\geq 0$$ which is true.

    Sonnhard.

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