Cyclic Doesn't Mean Symmetric!

$\frac{1+ab}{1+a} + \frac{1+bc}{1+b} + \frac{1+ca}{1+c}$

$=\frac{1+\frac{1}{c}}{1+a} + \frac{1+\frac{1}{a}}{1+b} + \frac{1+\frac{1}{b}}{1+c}$

$\Rightarrow \frac{c+1}{c(1+a)} + \frac{a+1}{a(1+b)} + \frac{b+1}{b(1+c)} \geq 3(\frac{c+1}{c(1+a)} \times \frac{a+1}{a(1+b)} \times \frac{b+1}{b(1+c)})^\frac{1}{3}$ [AM-GM inequality]

$\Rightarrow \frac{c+1}{c(1+a)} + \frac{a+1}{a(1+b)} + \frac{b+1}{b(1+c)} \geq 3(\frac{1}{abc})^\frac{1}{3}$

$\Rightarrow \frac{c+1}{c(1+a)} + \frac{a+1}{a(1+b)} + \frac{b+1}{b(1+c)} \geq \boxed{3}$

Since $a$ , $b$ and $c$ are positive reals, we can use AM-GM inequality as follows:

$\begin{aligned} \frac{1+ab}{1+a} + \frac{1+bc}{1+b} + \frac{1+ca}{1+c} & \ge 3 \sqrt[3]{\frac{\color{#3D99F6}{1}+ab}{1+a} \dot{} \frac{\color{#3D99F6}{1}+bc}{1+b} \dot{} \frac{\color{#3D99F6}{1}+ca}{1+c}} \quad \quad \small \color{#3D99F6}{\text{As } abc = 1} \\ & \ge 3 \sqrt[3]{\frac{\color{#3D99F6}{abc}+ab}{1+a} \dot{} \frac{\color{#3D99F6}{abc}+bc}{1+b} \dot{} \frac{\color{#3D99F6}{abc}+ca}{1+c}} \\ & \ge 3 \sqrt[3]{\frac{ab(c+1)}{1+a} \dot{} \frac{bc(a+1)}{1+b} \dot{} \frac{ca(b+1)}{1+c}} \\ & \ge 3 (abc)^{\frac{2}{3}} = \boxed{3} \end{aligned}$

A useful transformation for constraints like abc = 1 is a = x/y, b = y/z, c = z/x.

The first term, (1 + ab)/(1+a) = y(z+x)/z(y+x)

Since the R.H.S is cyclic in its original form, it's quite easy to see that the AM-GM inequality yields a 1 on the right side of the inequality.

Hence, the minimum value of the expression is 3.

$\dfrac{1+ab}{1+a}+\dfrac{1+bc}{1+b}+\dfrac{1+ca}{1+c}$

$=\dfrac{abc+ab}{1+a}+\dfrac{abc+bc}{1+b}+\dfrac{abc+ca}{1+c}$

$=\dfrac{ab(1+c)}{1+a}+\dfrac{bc(1+a)}{1+b}+\dfrac{ac(1+b)}{1+c}$

$\geq 3\sqrt[3]{\dfrac{ab(1+c)}{1+a}\dfrac{bc(1+a)}{1+b}\dfrac{ac(1+b)}{1+c}}=\boxed{3}$

on simplifying the given equation we get, numerator = 6+3a+3b+3c+2/a+2/b+2/c+a/b+b/c+c/a denominator = a+b+c+1/a+1/b+1/c+2 now, numerator = 1+1+1+1+1+1+a+a+a+b+b+b+c+c+c+1/a+1/a+1/b+1/b+1/c+1/c+a/b+b/c+c/a denominator = a+b+c+1/a+1/b+1/c+1+1 AM>=GM numerator n/24 >= 1 ("n" is numerator) this implies n>= 24 denominator d/8>=1 ("d" is denominator) this implies d>=8 THEREFORE, n/d is greater than or equal to 24/8 this implies n/d >= 3 HENCE proved that the least value of the given question is equal to 3 SORRY I DON'T KNOW HOW TO WRITE PROPERLY

$\frac{1+ab}{1+a}$ + $\frac{1+bc}{1+b}$ + $\frac{1+ca}{1+c}$

Given abc=1

=> $\frac{1+\frac{ab}{abc}}{1+a}$ + $\frac{1+\frac{bc}{abc}}{1+b}$ + $\frac{1+\frac{ca}{abc}}{1+c}$

=> $\frac{1+c}{c(1+a)}$ + $\frac{1+a}{a(1+b)}$ + $\frac{1+b}{b(1+c)}$ $\geq$ 3 $\sqrt[3]{ \frac{1}{abc} \frac{1+c}{1+a} \frac{1+c}{1+b} \frac{1+b}{1+c} }$ By AM-GM Inequality

Therefore $\frac{1+ab}{1+a}$ + $\frac{1+bc}{1+b}$ + $\frac{1+ca}{1+c}$ $\geq$ 3 $\sqrt[3]{1}$ $\geq$ $\fbox{3}$

Assume a, b, and c are 1. The expression has some kind of symmetry between them. 1 + 1 + 1 = 3.