Find the maximum of the value $F=x^3y+y^3z+z^3x$ let $x,y,z$ be real number.if $x+y+z=3$,show that
$$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$
and  the inequality $=$,then $x=?,y=?,z=?$
I can solve if add $x,y,z\ge 0$,also see:  Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.
But for real $x,y,z$ I can't solve it


A SOS (Sum of Squares) solution with computer:
Denote $K = \frac{9(63+5\sqrt{105})}{32}$.
We have
$$
 K\left(\frac{x + y + z}{3}\right)^4 - (x^3y + y^3z + z^3x)
 = \alpha_1 f_1^2  
 + \alpha_2 f_2^2
 + \alpha_3 f_3^2
$$
where
\begin{align*}
 \alpha_1 &= \frac{567 - 16K}{13778100} > 0, \\[5pt]
 \alpha_2 &= \frac{472K - 5607}{292287082500} > 0,\\[5pt]
 \alpha_3 &= \frac{998K - 29538}{1491260625} > 0, \\[8pt]
 f_1 &= \left( 10\,{x}^{2}+20\,xy+20\,xz-14\,{y}^{2}+44\,yz-14\,{z}^{2}
 \right) K \\
 &\qquad -405\,xy+459\,{y}^{2}-999\,yz+459\,{z}^{2},\\[8pt]
 f_2 &= \left( 944\,xy-112\,xz-592\,{y}^{2}+1440\,yz-480\,{z}^{2} \right) K \\ 
 &\qquad -
 22239\,xy+8442\,xz+21987\,{y}^{2}-40635\,yz+17325\,{z}^{2},\\[8pt]
 f_3 &= \left( 16\,{y}^{2}-48\,yz \right) K+45\,xz-261\,{y}^{2}+918\,yz-405\,
 {z}^{2}.
\end{align*}
