A Killer Inequality

$a+b+c=3\implies \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq 3$ . (A. M. -H. M. inequality).

Hence $\left (a+\dfrac{1}{c}\right )+\left (b+\dfrac{1}{a}\right )+\left (c+\dfrac{1}{b}\right )\geq 6$ ,

i. e. $\dfrac{ab+1}{a}+\dfrac {bc+1}{b}+\dfrac{ca+1}{b}\geq 6$ .

Now $\left (\dfrac{ab+1}{a}+\dfrac{bc+1}{b}+\dfrac{ca+1}{c}\right )\left (\dfrac{a}{ab+1}+\dfrac {b}{bc+1}+\dfrac{c}{ca+1}\right )\geq 9$ .

So, for the minimum of $\dfrac{ab+1}{a}+\dfrac{bc+1}{b}+\dfrac{ca+1}{c}$ ,

the minimum of $\dfrac{a}{ab+1}+\dfrac{b}{bc+1}+\dfrac{c}{ca+1}$ is $\dfrac{9}{6}=\boxed {1.5}$ .

Similar solution as @Alak Bhattacharya

$\begin{aligned} \frac a{ab+1} + \frac b{bc+1} + \frac c{ca+1} & = \blue{\frac 1{b+\frac 1a} + \frac 1{c+\frac 1b} + \frac 1{a+\frac 1c}} & \small \blue{\text{By Hölder's inequality}} \\ \left(\blue{\frac 1{b+\frac 1a} + \frac 1{c+\frac 1b} + \frac 1{a+\frac 1c}}\right)\left(b+ \red{\frac 1a} + c + \red{\frac 1b} + a + \red{\frac 1c} \right)(1+1+1) & \ge (1+1+1)^3 & \small \red{\text{By AM-HM inequality}} \\ \left(\frac a{ab+1} + \frac b{bc+1} + \frac c{ca+1}\right)\left(a+b+c + \red{\frac 9{a+b+c}} \right)(3) & \ge 27 \\ \implies \frac a{ab+1} + \frac b{bc+1} + \frac c{ca+1} & \ge \frac 32 = \boxed{1.5} \end{aligned}$

---

References:

• Hölder's inequality
• AM-HM inequality