A genius solution!

$\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2$

$=\left[\frac{a}{ab+a+1}+\frac{b(a)}{(bc+b+1)(a)}+\frac{c(ab)}{(ca+c+1)(ab)}\right]^2$

$=\left(\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{abc}{abca+abc+ab}\right)^2$

$=\left(\frac{a}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{1}{a+1+ab}\right)^2$

$=\left(\frac{ab+a+1}{ab+a+1}\right)^2$

$=1^2=\boxed{1}$

Another simplification is obtained by putting a = x/y, b = y/z and c = z/x, which gives same denominator, so that adding becomes easy..

Since they said that all the numbers are real values, as abc = 1, I can say that a = 1, b = 1, c = 1. As you sub in to the respective alphabeths, you will get, ((1/3) + (1/3) + (1/3))^2 = (1)^2 = 1

I noticed that, for example, $\frac{a}{a b+a+1} \\ = \frac{a}{a b+a+a b c} \\ = \frac{a}{a( b+1+bc)}\\=\frac{1}{b c +b+1}$ so there is a cyclic relation between the terms, which leads to making equal the numerator and the denominator.

Multiplying the top and bottom of $\frac{b}{bc+b+1}$ by a and the top and bottom of $\frac{c}{ca+c+1} )^2$ by ab, we get ( $\frac{a}{ab+a+1}$ + $\frac{b}{bc+b+1}$ + $\frac{c}{ca+c+1} )^2$ =( $\frac{a}{ab+a+1}$ + $\frac{ab}{abc+ab+a}$ + $\frac{abc}{a^2bc+abc+ab} )^2$ =( $\frac{ab+a+1}{ab+a+1} )^2$ =( $1 )^2$ =1

There's no need to assign any values for A , B & C , but we can solve for them as follows : $\frac{a}{ab+a+1}=\frac{a}{ab+a+abc}=\frac{1}{bc+b+1}$ and $\frac{b}{bc+b+1}=\frac{b}{bc+b+abc}=\frac{1}{ac+c+1}$ and $\frac{c}{ca+c+1}=\frac{c}{ca+c+abc}=\frac{1}{ab+a+1}$ and one finds from these equations that : $\frac{a}{ab+a+1}\equiv \left ( \frac{c}{ca+c+1} \right )a$ and $\frac{c}{ca+c+1}\equiv \left ( \frac{b}{bc+b+1} \right )c$ and $\frac{b}{bc+b+1}\equiv \left ( \frac{a}{ab+a+1} \right )b$ from these equations we can formulate the question in a different form : $\frac{ac}{ac+c+1}+\frac{ab}{ab+a+1}+\frac{bc}{bc+b+1}$ and we can try to unify all of these by multiplying each fraction by the term its missing : $\frac{abc}{abc+bc+b}+\frac{abc}{abc+ac+c}+\frac{abc}{abc+ab+a}$ and since abc =1 , the last equation becomes : $\frac{1}{1+bc+b}+\frac{1}{1+ac+c}+\frac{1}{1+ab+a}$ and from that we can equate the coefficients for A , B & C from the question itself to find that $A=1 , B=1 , C=1$

and from the last equation we can substitute the values we obtained to get : $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}= 1$ and by squaring it , we get one !

If the answer is the same regardless of the actual values of a, b, and c, make them all equal to 1, which satisfies abc = 1. Then the answer becomes very obvious.

A naive approach: Since $a$ , $b$ , and $c$ can be any real numbers for which we know that $\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2=constant$ ; therefore, we can simply take any random values for $a$ , $b$ , and $c$ , say $a=b=c=1$ , such that $abc=1$ and substitute in it as follows; $\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2=\left(\frac{1}{1+1+1}+\frac{1}{1+1+1}+\frac{1}{1+1+1}\right)^2=\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)^2=1$

I don't know if that's clear but here you go

$(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1})^2.$

$\Rightarrow (\frac{a}{ab+a+abc}+\frac{b}{bc+b+abc}+\frac{c}{ca+c+abc})^2$

$= ( \frac{1}{b+1+bc}+\frac{1}{c+1+ac}+\frac{1}{a+1+ab})^2$

$= ( \frac{1}{b+1+bc}+\frac{1}{c+1+ac}+\frac{1}{a+1+ab})^2$

$= (\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1})^2 \Leftrightarrow (\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1})^2$

The expression is cyclic. By comparing terms we get that

$(a,b,c) = (1,1,1)$

Hence the expression is equal to:

$\large (\frac{1}{3}+\frac{1}{3}+\frac{1}{3})^2 = \boxed{1}.$

For starter,

$a * b * c = 1$ therefore a = 1 & b = & c = 1

Therefore,

( $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ )² = 1

If abc = 1, then you know that a = 1, b = 1, and c = 1. Therefore, substitute 1 in the place of each variable. Once you do that, then you should have ( $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ )^2. Add the fractions, and you get $\frac{3}{3}$ or 1. 1^2 = 1.

Maybe I'm missing the point here, but just by looking at this one logic seems much quicker. If a,b,c are real numbers and abc=1 then we know all the numerators are 1. A quick glance at the denominators and they are all the same as well:
$(1*1)+(1)+1=3$
So now we've got...
$( \frac {1}{3} + \frac {1}{3} + \frac {1}{3} )^2 =(1)^2 =1$

$(\frac {a} {ab+a+1} + \frac {b} {bc+b+1}+ \frac {c} {ac+c+1})^2=(\frac {a} {ab+a+1} + \frac {b} {bc+b+1} \times \frac {a} {a}+ \frac {c} {ac+c+1} \times \frac {\frac {1} {c}} {\frac {1} {c}})^2$

= $(\frac {a} {ab+a+1} + \frac {ab} {ab+a+1}+ \frac {1} {ac+c+1})^2$

= $(\frac {ab+a+1} {ab+a+1})^2$

= $\boxed {1}$