<s>[INST] 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.
 [/INST] A NOTE
The given inequality
$$
\left(\sum_{cyc}\frac{a+b}{c}\right)^2\geq 4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}\frac{1}{a^2}\right)\textrm{, }a,b,c>0\tag 1
$$
is written as
$$
\left(\sum_{cyc}ab(a+b)\right)^2\geq4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2b^2\right).\tag 2
$$
Hence if $s=a+b+c$ and $p=abc$ and $s_h=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then
$$
\left(\sum_{cyc}ab(a+b)\right)^2=p^2(ss_h-3)^2
$$
and
$$
4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2b^2\right)=4p^3s_h(s_h^2-2s/p)
$$
Hence (2) becomes
$$
9+2ss_h+s^2s_h^2-4ps_h^3\geq 0.\tag 3
$$
Inequality (3) is a equivalent to
$$
9\leq ss_h\leq \rho\left(\frac{4p}{s^3}\right),
$$
where $\rho=\rho(t)$ is the real root of
$$
t \rho^3=9+2 \rho+\rho^2.
$$
Hence given positive numbers $a,b,c$ we have
$$
9\leq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\leq x,
$$
where $x$ is the real root of
$$
9+2x+x^2=\frac{4abc}{(a+b+c)^3}x^3.
$$
</s>