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 
 A: Note that
$$
\begin{align}\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}&=\left(\frac{x^2}{x+y}+y-x\right)+\left(\frac{y^2}{y+z}+z-y\right)+\left(\frac{z^2}{z+x}+x-z\right)
\\
&=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32\,.
\end{align}$$
Therefore,
$$\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}=3\,.$$
By the AM-GM Inequality, we see that
$$\left(\frac{x^2+y^2}{x+y}\right)\left(\frac{y^2+z^2}{y+z}\right)\left(\frac{z^2+x^2}{z+x}\right)\leq 1\tag{*}\,.$$
We claim that, for $a,b>0$,
$$\frac{a^3+b^3}{2}\leq \left(\frac{a^2+b^2}{a+b}\right)^3\,,\tag{#}$$
and the inequality becomes an equality if and only if $a=b$.  To justify the claim, let $p(t)\in\mathbb{R}[t]$ be the polynomial
$$p(t):=2\,\left(t^2+1\right)^3-(t+1)^3(t^3+1)\,.$$
Since $p(1)=16-16=0$, we know that $(t-1)$ is a factor of $p(t)$.  Now,
$$p'(t)=12\,t(t^2+1)^2-3\,(t+1)^2(t^3+1)-3\,t^2(t+1)^3\,,$$
which satisfies $p'(1)=48-24-24=0$ again.  That is, $(t-1)^2$ is a factor of $p(t)$.  We proceed further:
$$p''(t)=48\,t^2(t^2+1)+12\,(t^2+1)^2-6\,(t+1)(t^3+1)-18\,t^2(t+1)^2-6\,t(t+1)^3\,.$$
We have again that $p''(1)=96+48-24-72-48=0$, and so $(t-1)^3$ is a factor of $p(t)$.  Now,
$$p'''(t)=144\,t(t^2+1)+96\,t^3-6\,(t^3+1)-54\,t^2(t+1)-54\,t(t+1)^2-6\,(t+1)^3\,,$$
so $p'''(1)=288+96-12-108-216-48=0$, whence $(t-1)^4$ is a factor of $p(t)$.  Because $p$ is a monic polynomial of degree $6$, we must have
$$p(t)=(t-1)^4\,(t^2+\alpha t+\beta)$$
for some $\alpha,\beta\in\mathbb{R}$.  With $p(0)=1$, we get $\beta=1$.  As $p(-1)=16$, we conclude that $1-\alpha+\beta=1$, so that $\alpha=1$, as well.  Consequently,
$$p(t)=(t-1)^4\,(t^2+t+1)\,,$$
which is a nonnegative polynomial (i.e., $p(\mathbb{R})\subseteq \mathbb{R}_{\geq 0}$), and the only real root of $p(t)$ is $t=1$.
Now, (#) is equivalent to
$$2\,\left(a^2+b^2\right)^3-(a+b)^3\left(a^3+b^3\right)=b^6\,p\left(\frac{a}{b}\right)=(a-b)^4\,\left(a^2+ab+b^2\right)\geq 0\,,$$ 
which is an equality iff $a=b$.  Hence, the claim is established, but then we conclude that
$$\left(\frac{x^3+y^3}{2}\right)\left(\frac{y^3+z^3}{2}\right)\left(\frac{z^3+x^3}{2}\right)\leq 1\,,$$
using (#) in (*).  Therefore,
$$(x^3+y^3)(y^3+z^3)(z^3+x^3)\leq 8\,.$$
However, the problem statement demands that the inequality above is an equality.  That is, $x=y=z$ must hold.  Ergo, the only positive real solution to this system of equations is
$$(x,y,z)=(1,1,1)\,.$$
A: 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! 
