 # Solve in positive real numbers the system of equations: <mtable rowspacing="3pt" columnspacing= Raul Walker 2022-07-01 Answered
Solve in positive real numbers the system of equations:
$\begin{array}{c}\left({x}^{3}+{y}^{3}\right)\left({y}^{3}+{z}^{3}\right)\left({z}^{3}+{x}^{3}\right)=8,\\ \frac{{x}^{2}}{x+y}+\frac{{y}^{2}}{y+z}+\frac{{z}^{2}}{z+x}=\frac{3}{2}.\end{array}$
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Note that
$\begin{array}{rl}\frac{{y}^{2}}{x+y}+\frac{{z}^{2}}{y+z}+\frac{{x}^{2}}{z+x}& =\left(\frac{{x}^{2}}{x+y}+y-x\right)+\left(\frac{{y}^{2}}{y+z}+z-y\right)+\left(\frac{{z}^{2}}{z+x}+x-z\right)\\ & =\frac{{x}^{2}}{x+y}+\frac{{y}^{2}}{y+z}+\frac{{z}^{2}}{z+x}=\frac{3}{2}\phantom{\rule{thinmathspace}{0ex}}.\end{array}$
Therefore,
$\frac{{x}^{2}+{y}^{2}}{x+y}+\frac{{y}^{2}+{z}^{2}}{y+z}+\frac{{z}^{2}+{x}^{2}}{z+x}=3\phantom{\rule{thinmathspace}{0ex}}.$
By the AM-GM Inequality, we see that
$\begin{array}{}\text{(*)}& \left(\frac{{x}^{2}+{y}^{2}}{x+y}\right)\left(\frac{{y}^{2}+{z}^{2}}{y+z}\right)\left(\frac{{z}^{2}+{x}^{2}}{z+x}\right)\le 1\phantom{\rule{thinmathspace}{0ex}}.\end{array}$
We claim that, for $a,b>0$,
$\begin{array}{}\text{(#)}& \frac{{a}^{3}+{b}^{3}}{2}\le {\left(\frac{{a}^{2}+{b}^{2}}{a+b}\right)}^{3}\phantom{\rule{thinmathspace}{0ex}},\end{array}$
and the inequality becomes an equality if and only if $a=b$. To justify the claim, let $p\left(t\right)\in \mathbb{R}\left[t\right]$ be the polynomial
$p\left(t\right):=2\phantom{\rule{thinmathspace}{0ex}}{\left({t}^{2}+1\right)}^{3}-\left(t+1{\right)}^{3}\left({t}^{3}+1\right)\phantom{\rule{thinmathspace}{0ex}}.$
Since $p\left(1\right)=16-16=0$, we know that $\left(t-1\right)$ is a factor of $p\left(t\right)$. Now,
${p}^{\prime }\left(t\right)=12\phantom{\rule{thinmathspace}{0ex}}t\left({t}^{2}+1{\right)}^{2}-3\phantom{\rule{thinmathspace}{0ex}}\left(t+1{\right)}^{2}\left({t}^{3}+1\right)-3\phantom{\rule{thinmathspace}{0ex}}{t}^{2}\left(t+1{\right)}^{3}\phantom{\rule{thinmathspace}{0ex}},$
which satisfies ${p}^{\prime }\left(1\right)=48-24-24=0$ again. That is, $\left(t-1{\right)}^{2}$ is a factor of $p\left(t\right)$. We proceed further:
${p}^{″}\left(t\right)=48\phantom{\rule{thinmathspace}{0ex}}{t}^{2}\left({t}^{2}+1\right)+12\phantom{\rule{thinmathspace}{0ex}}\left({t}^{2}+1{\right)}^{2}-6\phantom{\rule{thinmathspace}{0ex}}\left(t+1\right)\left({t}^{3}+1\right)-18\phantom{\rule{thinmathspace}{0ex}}{t}^{2}\left(t+1{\right)}^{2}-6\phantom{\rule{thinmathspace}{0ex}}t\left(t+1{\right)}^{3}\phantom{\rule{thinmathspace}{0ex}}.$
We have again that ${p}^{″}\left(1\right)=96+48-24-72-48=0$, and so $\left(t-1{\right)}^{3}$ is a factor of $p\left(t\right)$. Now,
${p}^{‴}\left(t\right)=144\phantom{\rule{thinmathspace}{0ex}}t\left({t}^{2}+1\right)+96\phantom{\rule{thinmathspace}{0ex}}{t}^{3}-6\phantom{\rule{thinmathspace}{0ex}}\left({t}^{3}+1\right)-54\phantom{\rule{thinmathspace}{0ex}}{t}^{2}\left(t+1\right)-54\phantom{\rule{thinmathspace}{0ex}}t\left(t+1{\right)}^{2}-6\phantom{\rule{thinmathspace}{0ex}}\left(t+1{\right)}^{3}\phantom{\rule{thinmathspace}{0ex}},$
so ${p}^{‴}\left(1\right)=288+96-12-108-216-48=0$, whence $\left(t-1{\right)}^{4}$ is a factor of $p\left(t\right)$. Because $p$ is a monic polynomial of degree $6$, we must have
$p\left(t\right)=\left(t-1{\right)}^{4}\phantom{\rule{thinmathspace}{0ex}}\left({t}^{2}+\alpha t+\beta \right)$
for some $\alpha ,\beta \in \mathbb{R}$. With $p\left(0\right)=1$, we get $\beta =1$. As $p\left(-1\right)=16$, we conclude that $1-\alpha +\beta =1$, so that $\alpha =1$, as well. Consequently,
$p\left(t\right)=\left(t-1{\right)}^{4}\phantom{\rule{thinmathspace}{0ex}}\left({t}^{2}+t+1\right)\phantom{\rule{thinmathspace}{0ex}},$
which is a nonnegative polynomial (i.e., $p\left(\mathbb{R}\right)\subseteq {\mathbb{R}}_{\ge 0}$), and the only real root of $p\left(t\right)$ is $t=1$.
Now, (#) is equivalent to
$2\phantom{\rule{thinmathspace}{0ex}}{\left({a}^{2}+{b}^{2}\right)}^{3}-\left(a+b{\right)}^{3}\left({a}^{3}+{b}^{3}\right)={b}^{6}\phantom{\rule{thinmathspace}{0ex}}p\left(\frac{a}{b}\right)=\left(a-b{\right)}^{4}\phantom{\rule{thinmathspace}{0ex}}\left({a}^{2}+ab+{b}^{2}\right)\ge 0\phantom{\rule{thinmathspace}{0ex}},$
which is an equality iff $a=b$. Hence, the claim is established, but then we conclude that
$\left(\frac{{x}^{3}+{y}^{3}}{2}\right)\left(\frac{{y}^{3}+{z}^{3}}{2}\right)\left(\frac{{z}^{3}+{x}^{3}}{2}\right)\le 1\phantom{\rule{thinmathspace}{0ex}},$
using (#) in (*). Therefore,
$\left({x}^{3}+{y}^{3}\right)\left({y}^{3}+{z}^{3}\right)\left({z}^{3}+{x}^{3}\right)\le 8\phantom{\rule{thinmathspace}{0ex}}.$
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
$\left(x,y,z\right)=\left(1,1,1\right)\phantom{\rule{thinmathspace}{0ex}}.$

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