System of equations that can be solved by inequalities: $(x^3+y^3)(y^3+z^3)(z^3+x^3)=8$ and $\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32$ 
S367. Solve in positive real numbers the system of equations:
  \begin{gather*}
(x^3+y^3)(y^3+z^3)(z^3+x^3)=8,\\
\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32.
\end{gather*}
  Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

From https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2016-02/mr_2_2016_problems.pdf

I think I am smelling inequalities here. In the first equation I used Holder's inequality to show, $xyz \le 1$ , But in the second equation I used Titu's Lemma to get $x+y+z \le 3$ .But I think there would an equality case in one of the two equations. Can anyone help? The original source is Facebook 

The inequality $$\frac{a^2+b^2}{a+b}\geq\sqrt[3]{\frac{a^3+b^3}{2}}$$ we can prove also by AM-GM:
$$2(a^2+b^2)^3=\frac{2}{27}(3a^2+3b^2)^3=$$
$$=\frac{2}{27}\left(2(a^2-ab+b^2)+2\cdot\frac{(a+b)^2}{2}\right)^3\geq$$
$$\geq\frac{2}{27}\left(3\sqrt[3]{2(a^2-ab+b^2)\cdot\left(\frac{(a+b)^2}{2}\right)^2}\right)^3=$$
$$=(a^2-ab+b^2)(a+b)^4=(a^3+b^3)(a+b)^3$$
and we are done! 
