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 first inequality.
Let $a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $xy+xz+yz+xyz=4.$
Thus, the condition gives $$k^2(xy+xz+yz)+k^3xyz\leq xy+xz+yz+xyz$$ or
$$(k-1)((k+1)(xy+xz+yz)+(k^2+k+1)xyz)\leq0$$ or $$k\leq1.$$
Thus, we need to prove that $$8(x^2+y^2+z^2)\geq3k(x+y)(x+z)(y+z)$$ and since $0<k\leq1$,  it's enough to prove that $$8(x^2+y^2+z^2)\geq3(x+y)(x+z)(y+z).$$
Now, rewrite the new condition in the following form:
$$\sum_{cyc}\frac{1}{x+2}=4$$ and let $x=\frac{2p}{q+r}$ and $y=\frac{2q}{p+r},$ where $p$, $q$ and $r$ are positives.
Thus, $z=\frac{2r}{p+q}$ and after this substitution we obtain something obvious.
But it's better to prove before that $$x+y+z\geq xy+xz+yz,$$ for which we need to prove that:
$$\sum_{cyc}\frac{2p}{q+r}\geq\sum_{cyc}\frac{4pq}{(p+r)(q+r)}$$ or
$$\sum_{cyc}p(p+q)(p+r)\geq2\sum_{cyc}pq(p+q)$$ or
$$\sum_{cyc}(p^3-p^2q-p^2r+pqr)\geq0,$$ which is true by Schur.
Now, since $$1\geq\frac{xy+xz+yz}{x+y+z},$$ it's enough to prove that $$8(x^2+y^2+z^2)(xy+xz+yz)\geq3(x+y+z)(x+y)(x+z)(y+z)$$ or
$$\sum_{cyc}(5x^3y+5x^3z-6x^2y^2-4x^2yz)\geq0,$$ which is true by Muirhead.
