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.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, we need to prove that
$$\sum_{cyc}a^2b^2(a+b)^2\geq2a^2b^2c^2$$ or
$$\sum_{cyc}(a^4b^2+a^4c^2+2a^3b^3)\geq2a^2b^2c^2.$$
We see that $$\sum_{cyc}(a^4b^2+a^4c^2+2a^3b^3)\geq2a^2b^2c^2+\frac{1}{81}\left(\sum\limits_{cyc}(a^2b+a^2c-2abc)\right)^2$$
is a linear inequality of $w^3$, which says that it's enough to prove the last inequality
for an extremal value of $w^3$, which happens for equality case of two variables.
Since the last inequality is homogeneous and even degree, it's enough assume $b=c=1$,
which gives $$79a^4+170a^3-6a^2+7a+322\geq0,$$
which is obviously true.
Done!
