If $abc\neq 0$, then $ \frac{(a+b)^2}{c^2}+\frac{(a+c)^2}{b^2}+\frac{(b+c)^2}{a^2}\geq2 $ Let $a$, $b$ and $c$ be real numbers such that $abc\neq0$. Prove that:
$$ \frac{(a+b)^2}{c^2}+\frac{(a+c)^2}{b^2}+\frac{(b+c)^2}{a^2}\geq2 $$
I checked for some values and it seems to be true. But no plausible proof is there. I would love a counterexample, or a solution. But please no incomplete hints, I wish to see a complete solution, if it's true. Thanks.

Partial soln.
for $a,b,c \gt 0$ the value is same as for $a,b,c, \lt 0$. In both cases we get terms of the form $(\frac{a}{b})^2+(\frac{b}{a})^2+(\frac{a}{c})^2+(\frac{c}{a})^2+(\frac{c}{b})^2+(\frac{b}{c})^2$. This term $\ge 6$, and the remaining term is $2\left(\frac{ac}{b^2}+\frac{ba}{c^2}+\frac{cb}{a^2}\right)$ which is positive. So, when all are of same sign, total is more than 6.
In the cases of two negative and one positive number, or vice versa, let $a,b \gt 0$ and $c \lt 0$. Then the squared terms still give a minimum of 6. But the cross term becomes:$2\left(\frac{-ac}{b^2}+\frac{ba}{c^2}+\frac{-cb}{a^2}\right) = 2\left(\frac{a^3b^3+(-b^3c^3)+(-c^3a^3)}{(abc)^2}\right)$. The numnerator and denominator AM and GM of $a^3b^3$, $-c^3a^3$ and $-b^3c^3$. But it doesn't seem to be helping.
