$x,y,z > 0$, prove that $ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $ $x,y,z > 0$, prove that 
$$ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $$
Without using Schur's inequality,

Attempt:
By $C.S$:
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} = x^{3/2}y^{3/2} + y^{3/2}z^{3/2} + z^{3/2}x^{3/2} $$
$$ \le \sqrt{ x^{3} + y^{3} +z^{3}} \sqrt{ y^{3} + z^{3} + x^{3}} = x^{3} + y^{3} + z^{3} $$
then I have to prove 
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2}  \le 3xyz$$
this one is difficult. Another thing that we know by AM-GM: $ x^{3} + y^{3}+ z^{3} \ge 3xyz$.
 A: Let $x=e^a$, $y=e^b$, $z=e^c$ $\;$ 
and  $a\ge b \ge c$ (WLOG)
Our inequality is equivalent to 
$$3e^{a+b+c} + e^{3a}+ e^{3b} + e^{3c} \ge 2 \left( e^{3(a+c)/2} + e^{3(a+c)/2} + e^{3(a+b)/2} \right)$$ 
If $a+c\le 2b$ ,
This inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all $x\ge 0$ and $$ \left( \frac{3(a+b)}{2}, \frac{3(a+b)}{2}, \frac{3(a+c)}{2}, \frac{3(a+c)}{2}, \frac{3(b+c)}{2}, \frac{3(b+c)}{2} \right ) \prec  \left( 3a,3b,(a+b+c),(a+b+c),(a+b+c),3c \right)$$
The majorization holds because, 
$$3a\ge \frac{3(a+b)}{2}$$
$$3a+ 3b \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} $$
$$3a+ 3b + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$
$$3a+ 3b + (a+b+c) + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$
$$3a+ 3b + (a+b+c) + (a+b+c)+ (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}+ \frac{3(b+c)}{2}$$
$$3a+ \dots + 3b+ 3c = \frac{3(a+b)}{2}+ \dots + \frac{3(b+c)}{2} + \frac{3(b+c)}{2} $$
$3^{rd}$ and $4^{th}$ inequalities by $a+ c \le 2b$. $\;$ Other inequalities are obvious. 
If $a+c\ge 2b$ ,
The desired inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all $x\ge 0$ and $$ \left( \frac{3(a+b)}{2}, \frac{3(a+b)}{2}, \frac{3(a+c)}{2}, \frac{3(a+c)}{2}, \frac{3(b+c)}{2}, \frac{3(b+c)}{2} \right ) \prec  \left( 3a,(a+b+c),(a+b+c),(a+b+c),3b,3c \right)$$
The majorization holds because, 
$$3a\ge \frac{3(a+b)}{2}$$
$$3a+ (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} $$
$$3a+ (a+b+c) + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$
$\dots$
$$3a+ \dots + 3b+ 3c = \frac{3(a+b)}{2}+ \dots + \frac{3(b+c)}{2} + \frac{3(b+c)}{2} $$
$3^{rd}$ inequality comes from $a+ c \ge 2b$. $\;$ Other inequalities are obvious. 
A: First, Schur's inequality provides that
$$
x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x).
$$
Then, 
$$
xy(x+y)\ge 2xy\sqrt{xy} 
$$
and hence
$$
x^3+y^3+z^3+3xyz\ge 2\big(xy\sqrt{xy} +yz\sqrt{yz} +zx\sqrt{zx} \big)
$$
