inequality

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  • #70591
    Raghav Agrawal
    Participant

    proove a(a-1)+b(b-1)+c(c-1)>=0

    when (abc)=1

    #70870
    Saumik Karfa
    Moderator

    $\frac{a^{1+1}+b^{1+1}+c^{1+1}}{3}>\frac{a+b+c}{3}\cdot\frac{a+b+c}{3}\geq \sqrt[3]{abc}\frac{a+b+c}{3}=\frac{a+b+c}{3}[\text{ since }abc=1]$

    $\Rightarrow a^2+b^2+c^2\geq a+b+c$

    $\Rightarrow a(a-1)+b(b-1)+c(c-1)\geq 0$ [equality occurs when $a,b,c=1$]

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