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.

Assume that $a^2=\min\{a^2,b^2,c^2\}$. By the Cauchy-Schwarz inequality, we have
$$\left(\frac {c+a}{b}\right)^2+\left(\frac {a+b}{c}\right)^2\ge \frac{\left((c+a)+(-a-b)\right)^2}{b^2+c^2}=\frac{(b-c)^2}{b^2+c^2}.$$
On the other hand,
$$\left(\frac{b+c}{a}\right)^2> \frac{(b+c)^2}{b^2+c^2}.$$
Therefore,
$$\left(\frac{b+c}{a}\right)^2+\left(\frac {c+a}{b}\right)^2+\left(\frac {a+b}{c}\right)^2> \frac{(b+c)^2}{b^2+c^2}+\frac{(b-c)^2}{b^2+c^2}=2.$$
The equality does not hold.
Is this proof okay? I have another conjecture, which also seems to be true, under previous condition :
$$\frac{(a+b)^2}{c^2}+\frac{(a+c)^2}{b^2}+\frac{(b+c)^2}{a^2}\ge 2+
\frac{10(a+b+c)^2}{3(a^2+b^2+c^2)}$$ where equality holds at $a=b=c$ but no formal proof. Thanks.
