Fractions With Twists And Turns

Relevant wiki: Classical Inequalities - Problem Solving - Intermediate

$\sum^{a,b,c}_{\text{cyc}}\frac{a}{bc+1}=\sum^{a,b,c}_{\text{cyc}}\frac{a^2}{abc+a}\stackrel{\text{Titu's Lemma}}\geq\frac{(a+b+c)^2}{3abc+(a+b+c)}\\ \sum^{a,b,c}_{\text{cyc}}\frac{a}{bc+1}\geq \frac{3}{abc+1}$ Now: $\begin{aligned} \frac{a+b+c}{3}&\stackrel{\text{AM-GM}}\geq \sqrt[3]{abc}\\ \implies abc+1&\leq \left(\frac{a+b+c}{3}\right)^3+1\\ \frac{3}{abc+1}&\geq \frac{1}{\left(\frac{a+b+c}{3}\right)^3+1}\\ \implies \frac{3}{abc+1}&\geq \frac{3}{2}\end{aligned}$ Equality occurs when $a=b=c$

We can rewrite it as :

$\dfrac{a^2}{abc + a} + \dfrac{b^2}{abc + b} + \dfrac{c^2}{abc + c}$ Call it $S$ .

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By Titu's Lemma ,

$S \geq \dfrac{(a + b + c)^2}{3abc + a + b + c} = \dfrac{9}{3abc + 3} = \dfrac{3}{abc + 1}$ .

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By AM - GM Inequality , we have ,

$a + b + c \geq 3 \cdot \sqrt[3]{abc}$

$\Rightarrow 1 \geq abc \Rightarrow 2 \geq abc + 1 \Rightarrow \dfrac{2}{3} \geq \dfrac{abc + 1}{3}$

$\Rightarrow \dfrac{3}{2}\leq \dfrac{3}{abc + 1}$

Therefore , $S \geq \dfrac{3}{2}$

Equality holds when $\boxed{a = b = c = 1}$

Here's my U.C.T approach $\displaystyle\sum_{cyc}^{a,b,c} \frac{a}{bc+1}\stackrel{\text{AM-GM}}\geq\displaystyle\sum_{cyc}^{a,b,c}\frac{a}{\frac{(b+c)^2}{4}+1}=\displaystyle\sum_{cyc}^{a,b,c} \frac{4a}{(3-a)^2+4}=\displaystyle\sum_{cyc}^{a,b,c} \frac{4a}{a^2-6a+13}$ Now we must prove that $\frac{4a}{a^2-6a+13}\geq \frac{1}{2}+\frac{3}{2}(a-1)$ $\Leftrightarrow \frac{-(a-1)^2(3a-13)}{a^2-6a+13}\geq 0 \ ( True \ for \ every \ 0<x<3)$ Proving the similar for $\frac{4b}{b^2-6b+13}$ and $\frac{4c}{c^2-6c+13}$ then combine all 3 inequalities we get $\displaystyle\sum_{cyc}^{a,b,c} \frac{a}{bc+1}\geq\displaystyle\sum_{cyc}^{a,b,c} \frac{4a}{a^2-6a+13}\geq\frac{3}{2}$ So the minimum value is $\frac{3}{2}=1.50$ , the equality holds when $a=b=c=1$