<s>[INST] 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, 
$\frac{(x+2)^2}{3}(x+2\sqrt{x}+3) \geq 3\sqrt[3]{x(x^3+2)^2}$
$(x+2)^2(x+2\sqrt{x}+3) \geq 9\sqrt[3]{x(x^3+2)^2}$
 [/INST] 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 $t\in(0,1]$.
</s>