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.
 A: 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.
A: 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.
A: 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!
