Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$ 
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$

From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq\sqrt[3]x\left(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\right)$,
Prove that $(x+2)^2(x+2\sqrt{x}+3)\geq9\sqrt[3]{x(x^3+2)^2}$,
LHS :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq (x+2) \frac{x+2}
{\sqrt{3}}\left(\frac{x+2}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)= \frac{(x+2)^2}{3}(x+2\sqrt{x}+3)$
RHS :
$\sqrt[3]{3x(x^3+2)}\leq \sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{9x^2}\leq \sqrt[3]{(x^3+2)^2}$
so $\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\leq 3\sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{x}(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2})\leq 3\sqrt[3]{x(x^3+2)^2}$
Thus, 
$\frac{(x+2)^2}{3}(x+2\sqrt{x}+3) \geq 3\sqrt[3]{x(x^3+2)^2}$
$(x+2)^2(x+2\sqrt{x}+3) \geq 9\sqrt[3]{x(x^3+2)^2}$
 A: $$x+y+z=p,\ xy+yz+zx=q,\ xyz=r$$
we can assume :  $x+y+z=1$ 
$x^3+y^3+z^3=1-3q+3r,\ x^2+y^2+z^2=1-2q\ \ \ \left( 0 < q \le \dfrac{1}{3}\right)$
$$\left( \dfrac{1-3q}{r}+3\right)^{1/3}+\left( \dfrac{q}{1-2q}\right)^{1/2}\ge 1+\sqrt[3]3$$
We have
$$q^2=(xy+yz+zx)^2\ge 3xyz(x+y+z) =3r.$$
So: $LHS \ge \sqrt[3]3\cdot\left( \dfrac{1-3q}{q^2}+1\right)^{1/3}+\left( \dfrac{q}{1-2q}\right)^{1/2}=f(q)$
$$\dfrac{q}{1-2q}=t  \le 1$$
$g(t)= \sqrt[3]3\cdot\left( \dfrac{1+t}{t^2}-1\right)^{1/3}+ \sqrt{t}\ge  g\left(1\right )=1+\sqrt[3]3$
A: A proof by using the Sergic Primazon's idea. Dedicated to dear Hans.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that
$$\sqrt[3]{\frac{27u^3-27uv^2}{w^3}+3}+\sqrt{\frac{v^2}{3u^2-2v^2}}\geq\sqrt[3]3+1.$$
Now, since $27u^3-27uv^2\geq0$ and $v^4\geq uw^3$, it's enough to prove that
$$\sqrt[3]{\frac{27u^4-27u^2v^2}{v^4}+3}+\sqrt{\frac{v^2}{3u^2-2v^2}}\geq\sqrt[3]3+1.$$
Let $3u^2-2v^2=p^2v^2$, where $p>0$. 
Thus, $p\geq1$, $u^2=\frac{p^2+2}{3}v^2$ and we need to prove that
$$\sqrt[3]{3(p^2+2)^2-9(p^2+2)+3}+\frac{1}{p}\geq1+\sqrt[3]3$$ or
$$\sqrt[3]3\left(\sqrt[3]{p^4+p^2-1}-1\right)\geq\frac{p-1}{p}$$ or
$$\frac{\sqrt[3]{3}(p^4+p^2-2)}{\sqrt[3]{(p^4+p^2-1)^2}+\sqrt[3]{p^4+p^2-1}+1}\geq\frac{p-1}{p}$$ or
$$\sqrt[3]3p(p+1)(p^2+2)\geq\sqrt[3]{(p^4+p^2-1)^2}+\sqrt[3]{p^4+p^2-1}+1$$ and since
$$1\leq\sqrt[3]{p^4+p^2-1},$$ it's enough to prove that
$$\sqrt[3]3p(p+1)(p^2+2)\geq3\sqrt[3]{(p^4+p^2-1)^2}$$ or
$$p^3(p+1)^3(p^2+2)^3\geq9(p^4+p^2-1)^2$$ or
$$p^{12}+3p^{11}+9p^{10}+19p^9+21p^8+42p^7+26p^6+36p^5+33p^4+8p^3+18p^2-9\geq0.$$
Done!
A: Here is my, as Michael Rozenberg rightly dubs it, ugly proof of the last step in Sergic Primazon's proof that is $g(t)$ decreases for $t\in(0,1]$.
$$g'(t) = -3^{-\frac23}\Big(\frac1{t^2}+\frac1t-1\Big)^{-\frac23}\Big(\frac2{t^3}+\frac1{t^2}\Big)+\frac12\Big(\frac1t\Big)^{\frac12}\ .$$
We want to show $g'(t)<0$ or 
$$3^{\frac23}\Big(\frac1t\Big)^{\frac12}\Big(\frac1{t^2}+\frac1t-1\Big)^{\frac23} < 2\Big(\frac2{t^3}+\frac1{t^2}\Big). $$
Let $s:=\frac1t-1\in[0,\infty)$. The above inequality is equivalent to it being raised to the power of $6$, because it is easy to see its both sides are positive after making the substitution of $t$ by $s$. The transformed inequality is
$$2^6(s+1)^9(2s+3)^6-3^4((1+s)^2+s)^4>0 \tag1$$
The left hand side of the above inequality is
$$4096 s^{15} + 73728 s^{14} + 617472 s^{13} + 3191808 s^{12} + 11388672 s^{11} + 29712384 s^{10} + 58555200 s^9 + 88764399 s^8 + 104358708 s^7 + 95156902 s^6 + 66743280 s^5 + 35363277 s^4 + 13701744 s^3 + 3665574 s^2 + 605556 s + 46575 $$
All the coefficients of the polynomial are positive, so Inequality (1) holds for $s\ge 0$ implying $g'(t)<0$ for $t\in(0,1]$.
A: Denote $u = \frac{x^3+y^3+z^3}{3xyz}$ and $v = \frac{xy+yz+zx}{x^2+y^2+z^2}$.
Then $u\ge 1$ and $0 \le v \le 1$.
We need to prove that $\sqrt[3]{3u} + \sqrt{v} \ge 1 + \sqrt[3]{3}$ or
$\sqrt[3]{3u} - \sqrt[3]{3} \ge 1 - \sqrt{v}$ or
$$\frac{3u - 3}{3^{2/3}(u^{2/3} + u^{1/3} + 1)} \ge \frac{1 - v}{1 + \sqrt{v}}.$$
By using $(w^3+1) - (w^2+w) = (w+1)(w-1)^2\ge 0$ for $w\ge 0$, we have
$u^{2/3} + u^{1/3} \le u + 1$ (simply letting $w = u^{1/3}$). Also, $3^{2/3} < 3$ and $\frac{1 - v}{1 + \sqrt{v}} \le 1 - v$.
It suffices to prove that
$\frac{3u-3}{3(u+2)} \ge 1 - v$
or $\frac{u-1}{u+2}\ge 1 - v$ or
$$\frac{x^3+y^3+z^3 - 3xyz}{x^3+y^3+z^3+6xyz} \ge \frac{x^2+y^2+z^2-xy-yz-zx}{x^2+y^2+z^2}.$$
Since $x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - yz - zx)$, the above inequality is written as
$$\frac{(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2 - 6xyz)(x^2+y^2+z^2-xy-yz-zx)}{(x^3+y^3+z^3+6xyz)(x^2+y^2+z^2)} \ge 0.$$
By AM-GM, this inequality is true. We are done.
