For the **first inequality**, let us write $x=X^3$, $y=Y^3$, $z=Z^3$, and pass to a homogenized version of it, obtained by multiplying the R.H.S. with $(xyz)^{7/3}=(XYZ)^7$. So we have to show:
$$
(X^{30}+Y^{30}+Z^{30})^2 \ge 3 (X^{39}+Y^{39}+Z^{39})X^7Y^7Z^7\ .
$$
Now divide by $Z^{60}$ to dehomogenize, so it is enough to show the above for $Z=1$ and arbitrary $X,Y\ge 0$. Consider the difference function
$$
f(X,Y)=
(X^{30}+Y^{30}+1)^2 - 3 (X^{39}+Y^{39}+1)X^7Y^7\ .
$$
Since $f(1,1)=0$ and on the boundary $XY=0$ we have $f(X,Y)\ge 1$,
a global minimal value is a local minimum in the domain $Y,Y>0$. We search such local minima, they are among the solutions of $f'(X,Y)=0$, which leads to
the algebraic system of equations
$$
\left\{
\begin{aligned}
2(X^{30}+Y^{30}+1)\cdot 30X^{29}
&= 3 (46X^{45}Y^7 + 7X^6Y^{46}+7X^6Y^7)\ ,\\
2(X^{30}+Y^{30}+1)\cdot 30Y^{29}
&= 3 (46Y^{45}X^7 + 7Y^6X^{46}+7Y^6X^7)\ .
\end{aligned}
\right.
$$
Now multiply the first equation with $X$, the second with $Y$. This leads to
$$
2(X^{30}+Y^{30}+1)\cdot 30X^{30}
= 3 (46X^{46}Y^7 + 7X^7Y^{46}+7X^7Y^7)
= 2(X^{30}+Y^{30}+1)\cdot 30Y^{30}
\ .
$$
So a solution must satisfy $X=Y$. We plug in this into the above, get
$$
2(2X^{30}+1)\cdot 30X^{30}
= 3 (53X^{53}+7X^{14})
\ .
$$
It turns out that $X=1$ is the only real positive root. So $(X,Y)=(1,1)$ is the only critical point. So it is the point where $f$ is globaly minimal.

The **second point** follows from the inequality
$$
(2x^4+3y^4)(2y^4+3z^4)(2z^4+3x^4) \ge (3x+2y)(3y+2z)(3z+2x)x^3y^3z^3\ .
$$
Let us show the above. After we expand, it is enough to show the domination
$$
12 \, \sum x^{8} y^{4} + 18 \sum \, y^{8} x^{4}
\ge
18 \, \sum x^{5} y^{4} z^{3} + 12 \sum\, x^{5} y^{3} z^{4} \ .
$$
To have a better view, let us place the points $(i,j,k)$ corresponding to the monomials $x^iy^jz^k$ that occur in the plane $i+j+k=12$.

Now, a simple human scheme of domination can be found, for instance exploiting:
$$
\begin{aligned}
12(5,3,4) &= 5(8,4,0)+ 2(0,8,4) + 5(4,0,8)\ ,\\
12(5,4,3) &= 5(4,8,0)+ 2(0,4,8) + 5(8,0,4)\ .
\end{aligned}
$$
Now using the Jensen-convexity of the logarithm in the form $au+bv+cw\ge u^av^bw^c$ for $u,v,w>0$ and weights $a,b,c>0$ with total sum $a+b+c=1$, for the above suggested weights $a=c=5/12$, $b=2/12$, and the monomials $u=x^8y^4$, $v=y^8z^4$, $w=z^8x^4$ for the first line, then the reversed version for the second line, gives
$$
\begin{aligned}
12\,x^5y^3z^4 &\le 5\,x^8y^4 + 2\,y^8z^4 + 5\,z^8x^4\ ,\\
12\,x^5y^4z^3 &\le 5\,x^4y^8 + 2\,y^4z^8 + 5\,z^4x^8\ .
\end{aligned}
$$

This concludes the needed domination.