Let $m=(abc)^{\frac{1}{3}}$, where $a,b,c \in \mathbb{R^{+}}$. Then prove that

$\frac{b}{ab+b+1} + \frac{c}{bc+c+1} + \frac{a}{ac+a+1} \ge \frac{3m}{m^2+m+1}$

In this inequality I first applied Titu's lemma ; then Rhs will come 9/(some terms) ; now to maximise the rhs I tried to minimise the denominator by applying AM-GM inequality.But then the reverse inequality is coming Please help.


By Holder: $$\sum_{cyc}\frac{b}{ab+b+1}=1-\frac{(abc-1)^2}{\prod\limits_{cyc}(ab+b+1)}\geq1-\frac{(m^3-1)^2}{(\sqrt[3]{a^2b^2c^2}+\sqrt[3]{abc}+1)^3}=$$ $$=1-\frac{(m-1)^2(m^2+m+1)^2}{(m^2+m+1)^3}=1-\frac{(m-1)^2}{m^2+m+1}=\frac{3m}{m^2+m+1}.$$ Done!

  • 3
    $\begingroup$ Michael you are a legend! $\endgroup$ – Legend Killer May 3 '17 at 15:05

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