Just an interesting question I saw online. :)
Is the following statement true or false?
$$\large \dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} = 3 \iff \dfrac{a + b}{c} + \dfrac{b + c}{a} + \dfrac{c + a}{b} \in \{-3; 6\}$$
$a + b + c \ne 0$
Edit: In case you don't know, $a, b, c$ can be negative numbers too.
On
If $a,b,c$ are all positive then we have by Am-Gm:
$$\dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} \geq 3\sqrt[3]{\dfrac{ab}{c^2} \cdot \dfrac{bc}{a^2} \cdot \dfrac{ca}{b^2}} = 3$$
With eqaulity iff $ \dfrac{ab}{c^2} = \dfrac{bc}{a^2} = \dfrac{ca}{b^2}$ which is iff $a=b=c$ so the value of expression is $6$.
Now suppose not all are positive. Clearly not all numbers $a,b,c$ can be negative. Suppose $a$ is positive and $b$ negative.
Put $x=a/c$ and $y=b/c$, then we can rewrite starting equation like $$x^3y^3+x^3+y^3=3x^2y^2$$
If we put $m= xy<0$ and $k=x+y$ we get $$ m^3+k^3-3mk-3m^2=0\implies k=-m \;\;\;{\rm or} \;\;\;k^2-km+m^2=3m$$
We would like to know what is the value of $${(x+y)(xy+x+y+1)\over xy}-2$$ i.e. $$E:={k(m+k+1)\over m}-2$$
If $m=-k$ we get $E=-3$.
If $3m =k^2-km+m^2={k^2+m^2+(k-m)^2\over 2}\geq 0$, then $m\geq 0$ but this is a contradiction since $m<0$.
Notice that the equality $\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0\leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 \hspace{0.3cm}(1)$ or $a=b=c \hspace{0.3cm}(2)$.
We easily find by applying (1) solved for $a+b, b+c, c+a $ respectively:
$$ \begin{align} &\frac{a+b}{c}=-\frac{ab}{c^2}\\ &\frac{c+b}{a}=-\frac{bc}{a^2}\\ &\frac{a+c}{b}=-\frac{ac}{b^2} \end{align} $$ and we finally find that $$\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}=-\frac{ab}{c^2} -\frac{bc}{a^2} -\frac{ac}{b^2}=-3$$
by the original equation.
$$\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}=6$$
That means that the values of the algebraic expression in question could be either -3 or 6 given the constraint above.