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?
 A: 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.
A: 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 $a=b=c=1$. $\square$
A: 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.
