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

Let $a+b+c=x(a^2+b^2+c^2).$

Thus, $$\sum_{cyc}(a-xa^2)^2\geq0=\left(\sum_{cyc}(a-xa^2)\right)^2$$ or $$\sum_{cyc}(a^2-2xa^3+x^2a^4)\geq\sum_{cyc}(a^2+2ab)-2x\sum_{cyc}(a^3+a^2b+a^2c)+x^2\sum_{cyc}(a^4+2a^2b^2)$$ or $$\sum_{cyc}ab-x\sum_{cyc}(a^2b+a^2c)+x^2\sum_{cyc}a^2b^2\leq0.$$ Now, let $a^2b^2+a^2c^2+b^2c^2=A$, $\sum\limits_{cyc}ab(a+b)=B$ and $ab+ac+bc=C$.

We proved that there is $x$, id est, $x=\frac{a+b+c}{a^2+b^2+c^2}$ for which $A>0$ and $$Ax^2-Bx+C\leq0,$$ which gives $$\Delta\geq0$$ or $$B^2-4AC\geq0$$ or $$\left(\sum_{cyc}(a^2b+a^2c)\right)^2-4\sum_{cyc}ab\sum_{cyc}a^2b^2\geq0,$$ which is your inequality exactly.

WLOG assume $c\neq \text{mid}\{a,b,c\}$. We have:

$$a^2 b^2 c^2 (\text{LHS-RHS}) =\left( a-b \right) ^{2} \left( ab+ca+bc-{c}^{2} \right) ^{2}+4\,ab{c} ^{2} \left( a-c \right) \left( b-c \right) \geqq 0$$ Done.