Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$. 
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$

Let $ca + ab = m$, $ab + bc = n$ and $bc + ca = p$, we have that $$\left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right)^2 \ge 2(m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
$$\iff \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2} - 1\right)^2 \ge 2 \cdot \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right] + 1$$
Expanding $\displaystyle \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right]$ gives $$2 \cdot \sum_{cyc}\frac{ca}{b^2} + \left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)$$
Let $\dfrac{b + c}{a} = m'$, $\dfrac{c + a}{b} = n'$ and $\dfrac{a + b}{c} = p'$, we have that $$(m' + n' + p' - 1)^2 \ge 2 \cdot \left[2 \cdot \sum_{cyc}\frac{ca}{b^2} + (m' + n' + p')\right] + 1$$
Moreover, $$(m')^2 + (n')^2 + (p')^2 = \sum_{cyc}\left[\left(\frac{c + a}{b}\right)^2\right] \ge 2 \cdot \sum_{cyc}\frac{ca}{b^2}$$
$$\implies (m' + n' + p' - 1)^2 \ge 2 \cdot \left[(m')^2 + (n')^2 + (p')^2 + m' + n' + p'\right] + 1$$
$$\iff -[(m')^2 + (n')^2 + (p')^2] + 2(m'n' + n'p' + p'm') - 4(m' + n' + p') \ge 0$$, which is definitely not correct.
Another attempt, let $(0 <) \ a \le b \le c \implies ab \le ca \le bc \iff ca + ab \le ab + bc \le bc + ca$
$\iff m \le n \le p$ and $a^2 \le b^2 \le c^2 \iff \dfrac{1}{a^2} \ge \dfrac{1}{b^2} \ge \dfrac{1}{c^2}$.
By the Chebyshev inequality, we have that $$3 \cdot \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right) \le (m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Any help would be appreciated.

We use the standard pqr method.
Let $p = a + b + c, q = ab+bc+ca, r = abc$. The inequality becomes
$$\left(\frac{pq}{r} - 3\right)^2 \ge 4q \left(\left(\frac{q}{r}\right)^2 - 2\frac{p}{r}\right)$$
or (after clearing the denominators)
$$p^2q^2 + 2pqr - 4q^3 + 9r^2 \ge 0.$$
We split into two cases:
1) $p^2 \ge 4q$: Since $p^2q^2 \ge 4q^3$, the inequality is true.
2) $p^2 < 4q$: By Schur's inequality $a^2(a-b)(a-c) + b^2(b-c)(b-a) + c^2(c-a)(c-b) \ge 0$
which is written as $6pr - (4q-p^2)(p^2-q) \ge 0$, we have $r \ge \frac{(4q-p^2)(p^2-q)}{6p}$.
As $p^2 \ge 3q$, we have $r \ge \frac{(4q-p^2)(p^2-q)}{6p}\ge 0$.
Thus, we have 
\begin{align}
&p^2q^2 + 2pqr - 4q^3 + 9r^2\\
\ge\ & p^2q^2 + 2pq \cdot \frac{(4q-p^2)(p^2-q)}{6p} - 4q^3 + 9\left(\frac{(4q-p^2)(p^2-q)}{6p}\right)^2\\
= \ & \frac{(p^2-3q)(3p^2-q)(p^2-4q)^2}{12p^2}\\
\ge \ & 0.
\end{align}
We are done.
