Inequality (b)/(ab+b+1)+(c)/(bc+c+1)+(a)/(ac+a+1)>=(3m)/(m^2+m+1) Let m=(abc)^(1/3), where a,b,c in RR^+. 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.

Orion Cervantes 2022-09-25 Answered
Inequality $\frac{b}{ab+b+1}+\frac{c}{bc+c+1}+\frac{a}{ac+a+1}\ge \frac{3m}{{m}^{2}+m+1}$
Let $m=\left(abc{\right)}^{\frac{1}{3}}$, where $a,b,c\in {\mathbb{R}}^{\mathbb{+}}$. 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.
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Answers (1)

edytorialkp
Answered 2022-09-26 Author has 10 answers
By Holder:
$\sum _{cyc}\frac{b}{ab+b+1}=1-\frac{\left(abc-1{\right)}^{2}}{\prod _{cyc}\left(ab+b+1\right)}\ge 1-\frac{\left({m}^{3}-1{\right)}^{2}}{\left(\sqrt[3]{{a}^{2}{b}^{2}{c}^{2}}+\sqrt[3]{abc}+1{\right)}^{3}}=$
$=1-\frac{\left(m-1{\right)}^{2}\left({m}^{2}+m+1{\right)}^{2}}{\left({m}^{2}+m+1{\right)}^{3}}=1-\frac{\left(m-1{\right)}^{2}}{{m}^{2}+m+1}=\frac{3m}{{m}^{2}+m+1}.$
Done!
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