<s>[INST] Min $P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1}$ 
Given $x,y,z$ are  positive numbers such that $$x^2y+y^2z+z^2x=3$$
Find the minium of value: $$P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1} $$

My Attempt:
$$P= x^3y+y^3z+z^3x - \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1}  \right)$$
$$x^2+1 \geq 2x \Rightarrow \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1}\right)\leq \frac{1}{2}\left(x^2y+y^2z+z^2x\right)=\frac{3}{2}$$
$$P \geq x^3y+y^3z+z^3x - \frac{3}{2}$$
I was stuck when I was finding the minium of $x^3y+y^3z+z^3x$

Help me, thank you so much
 [/INST] There's no minimum of $P$ at all. Consider an intersection of $x^2y+y^2z+z^2x = 3$ and $y = x^{-4}$, i.e. $x^{-2} + x^{-8}z + z^2x = 3$ or $x^9 z^2 + z + x^6 - 3x^8 = 0$. We can find from here that there exists a positive solution
$$
z = \dfrac{-1 + \sqrt{1+12x^{17}-4x^{15}}}{2x^9}.
$$
at least for $x$ big enough. Moreover,
$$
\lim_{x \to +\infty}z\sqrt{x} = \sqrt{3}.
$$
Hence
$$
\lim_{x \to +\infty}\left[ \dfrac{x^5 y}{x^2+1} + \dfrac{y^5 z}{y^2+1} + \dfrac{z^5 x}{z^2+1}\right] = \lim_{x \to +\infty}\left[\dfrac{y^5 z}{y^2+1} + \dfrac{z^5 x}{z^2+1}\right] = \lim_{x \to +\infty}\dfrac{z^5 x}{z^2+1} = \lim_{x \to +\infty}\dfrac{3^{\frac{5}{2}}x^{-\frac{5}{2}} x}{3 x^{-1}+1} = 0.
$$
So $P$ does not have a minimal value on the set $(x,y,z) \in (0, +\infty) \times (0, +\infty) \times (0, +\infty)$ (because $P$ is obviously positive on it).
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