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?

This is an approach using Lagrange multipliers.
For the first part, we can write the problem as \begin{align}\min&\quad8(a^2+b^2+c^2)-3(a+b)(a+c)(b+c)\\\text{s.t.}&\quad ab+ac+bc+abc=4-\epsilon\\&\quad a,b,c>0\quad\land\quad0\le\epsilon<4.\end{align} Then we have $\mathcal L=f-\lambda g$ where $f(a,b,c)=8(a^2+b^2+c^2)-3(a+b)(a+c)(b+c)$ and $g(a,b,c)=ab+ac+bc+abc-(4-\epsilon)$. The partial derivatives are \begin{align}\mathcal L_a&=16a-3(b+c)(2a+b+c)-\lambda(b+c+bc)\\\mathcal L_b&=16b-3(a+c)(2b+a+c)-\lambda(a+c+ac)\\\mathcal L_c&=16c-3(a+b)(2c+a+b)-\lambda(a+b+ab)\\\mathcal L_\lambda&=ab+ac+bc+abc-(4-\epsilon).\end{align} Next, we have \begin{align}\mathcal L_a-\mathcal L_b&=0\implies16+3(a+b)+\lambda(1+c)=0,a=b\\\mathcal L_a-\mathcal L_c&=0\implies16+3(a+c)+\lambda(1+b)=0,a=c\\\mathcal L_b-\mathcal L_c&=0\implies16+3(b+c)+\lambda(1+a)=0,b=c,\end{align} so without loss of generality we have $a=b$. Letting $c=ka$ yields $f(a,b,c)=8(2+k^2)a^2-6(1+k)^2a^3$ and $g(a,b,c)=(1+2k)a^2+ka^3-(4-\epsilon)$. Since $f\to0^+$ as $\epsilon\to4^-$ we aim to find $k,\epsilon$ such that $f\le0$.
Elementary calculus reveals that $f(a;k)$ increases monotonically in the interval $[0,k^*]$ where $k^*=8(2+k^2)/(9(1+k)^2)$, from $0$ to $f(k^*;k)>0$. For $a>k^*$, the function $f(a;k)$ decreases monotonically to $-\infty$, where it meets the axis at $a=3k^*/2$. Notice that positive root of $g$ is largest when $4-\epsilon$ is greatest; that is, $\epsilon=0$. At this value, it suffices to notice that $$g\left(\frac{3k^*}2;k\right)=(1+2k)\left(\frac{4(2+k^2)}{3(1+k)^2}\right)^2+k\left(\frac{4(2+k^2)}{3(1+k)^2}\right)^3-4=(k-1)^2\cdot\frac{P(k)}{Q(k)}$$ where $P,Q$ are polynomials with positive coefficients. As $k>0$ the only solution to $g=0$ is $k=1$, from which it follows that $a=b=c=1$. $\square$
