Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$ 
Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$.  Prove the following inequalities:
(a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and
(b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\dfrac{2}{b^2c}+\dfrac{2}{c^2a}\geq 9$.
Prove also that the unique equality case for both inequalities is given by $a=b=c=1$.

Below are some probably useful or relevant results.

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*https://artofproblemsolving.com/community/c6h1241430p6342224

*https://artofproblemsolving.com/community/c6h284290p1535893

*https://artofproblemsolving.com/community/c6h608971p3619202

*https://artofproblemsolving.com/community/c6h1804479p11995588

*If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$
Techniques used in solving the inequalities in these links may prove useful in proving our inequalities.
Attempt.  In the simplest case, $a=b=c=:t$, we have $t^3+3t^2-4\leq 0$, whence $0<t\leq 1$.  Therefore, the inequalities (a) and (b) become
$$24t^2\geq 24t^3$$
and
$$\frac{3}{t^2}+\frac{6}{t^3}\geq 9\,,$$
which are obviously true.  How to prove these inequalities in general?

The source of the problem is: https://math.stackexchange.com/questions/2825783/problems-regarding-inequality.  Below are solutions for both parts, some were written by other users in the old thread above.
Part (a): Solution by Truth.
Let $a = \dfrac{2kx}{y+z}$, $b = \dfrac{2ky}{z+x}$, and $c = \dfrac{2kz}{x+y}$, where $x,y,z,k>0$.    From the given condition, we get $k \leqslant  1$ and the required inequality is equivalent to $$96k^2\,\sum_\text{cyc}\, \frac{x^2}{(y+z)^2} \geqslant 72k^3 \prod_\text{cyc}\, \left(\frac{x}{y+z}+\frac{y}{z+x}\right)\,.$$
This is equivalent to $$4\,\sum_\text{cyc}\, \frac{x^2}{(y+z)^2} \geqslant 3k\,\prod_\text{cyc}\, \left(\frac{x}{y+z}+\frac{y}{z+x}\right)\,.$$
Because $ k \leqslant 1$, it suffices to show that $$4\,\sum_\text{cyc}\, \frac{x^2}{(y+z)^2} \geqslant 3\, \prod_\text{cyc}\, \left(\frac{x}{y+z}+\frac{y}{z+x}\right)\,,$$ However, the last inequality is equivalent to
$$\sum_\text{cyc}\, \frac{\big(2x^4+16xyz^2+6y^2z^2+(2x^2+9xz+9yz+6z^2)(x+y-z)^2\big)(x-y)^2}{(x+y)^2(y+z)^2(z+x)^2} \geqslant 0\,.$$
Part (b): Solution by Michael Rozenberg.  (@Michael, if you want to use this proof in your own separate answer, then you can remove this solution from my answer here and add it into your own answer.)
By the AM-GM Inequality,
$$4\geq3\sqrt[3]{a^2b^2c^2}+abc\,,$$ which gives $$abc\leq1\,.$$
Thus, by the AM-GM Inequality again, we obtain
$$\sum_\text{cyc}\,\frac{1}{a^2}+2\,\sum_\text{cyc}\,\frac{1}{a^2b}\geq\frac{3}{\sqrt[3]{a^2b^2c^2}}+\frac{6}{abc}\geq9\,.$$
Part (b):  Alternative Proof by Me.
Use the AM-GM Inequality with the constraint inequality, we have
$$4\geq bc+ca+ab+abc\geq 4\,\sqrt[4]{(bc)(ca)(ab)(abc)}=4\,(abc)^{\frac{3}{4}}\,,$$
whence
$$abc\leq 1\,.$$
Now, we have by the AM-GM Inequality that
$$\sum_\text{cyc}\,\frac{1}{a^2}+2\,\sum_\text{cyc}\,\frac{1}{a^2b}\geq 9\, \sqrt[9]{\left(\prod_\text{cyc}\,\frac{1}{a^2}\right)\,\left(\prod_\text{cyc}\,\frac{1}{a^2b}\right)^2}=\frac{9}{(abc)^{\frac{8}{9}}}\,.$$
As $abc\leq 1$, the required inequality follows immediately.
