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: We can use the $uvw$'s technique here.
Indeed, let $\dfrac{9(63+5\sqrt{105})}{32}=k$, $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Thus, we need to prove that:
$$ku^4\geq\sum_{cyc}x^3y$$ or
$$2ku^4-\sum_{cyc}(x^3y+x^3z)\geq\sum_{cyc}(x^3y-x^3z)$$ or
$$2ku^4-(27u^2v^2-18v^4-3uw^3)\geq3u(x-y)(x-z)(y-z).$$
We'll prove that $2ku^4-(27u^2v^2-18v^4-3uw^3)\geq0.$
Indeed, this inequality is a linear inequality of $w^3$, which by $uvw$ says that it's enough to prove this inequality for equality case of two variables.
Let $y=z=1$.
We need to prove that $$\frac{2k}{81}(x+2)^4-2x^3-2x-2\geq0$$ and since $2k>27$, it's enough to prove that
$$(x+2)^4\geq6(x^3+x+1)$$ or $$x^4+2x^3+24x^2+26x+10\geq0$$ or
$$x^2(x+1)^2+23x^2+26x+10\geq0,$$ which is obvious.
Id est, it's enough to prove that:
$$(2ku^4-(27u^2v^2-18v^4-3uw^3))^2\geq9u^2\prod_{cyc}(a-b)^2$$ or
$$(2ku^4-(27u^2v^2-18v^4-3uw^3))^2\geq243u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6),$$
which is a quadratic inequality of $w^3$.
Now, prove that $\Delta\leq0.$
Can you end it now?
A: It is not surprizing that Mathematica does not produce explicit expressions for $(x,y,z)$.
Using it with the restriction $x>y>z$, it gives that $y$ and $z$ are respectively  the fourth and the first roots of the equation
$$56 t^6-336 t^5+378 t^4+819 t^3-2079 t^2+1512 t-351=0$$
Using  @River Li's comment, $(x,y,z)$ are the solutions of the cubic
$$u^3-3u^2-\frac{3}{8} \left(3+\sqrt{105}\right)u+\frac{3}{112} \left(147+17 \sqrt{105}\right)=0$$ and, using the trigonometric method, they are
$$x=1+\sqrt{\frac{11+\sqrt{105}}{2} } \cos \left(\frac{1}{3} \cos
   ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
   \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=3.71382\cdots$$
$$y=1-\sqrt{\frac{11+\sqrt{105}}{2} } \sin \left(\frac{\pi
   }{6}-\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
   \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=1.20642\cdots$$
$$z=1-\sqrt{\frac{11+\sqrt{105}}{2} } \sin \left(\frac{\pi
   }{6}+\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
   \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=-1.92024\cdots$$
Trying to convert to radicals would be more than problematic.
Edit
After a series of mistakes, using @River Li's comments, the factorization of the sextic as the product of two cubic polynomials
$$(t^3 + b_2t^2 + b_1t + b_0)\,(t^3 + a_2t^2 + a_1t + a_0)$$gives
$$b_2=-3 \qquad \qquad b_1=-\frac{3}{8} \left(3-\sqrt{105}\right)\qquad \qquad b_0=\frac{3}{112} \left(147-17 \sqrt{105}\right)$$
$$a_2=-3 \qquad \qquad a_1=-\frac{3}{8} \left(3+\sqrt{105}\right)\qquad \qquad a_0=\frac{3}{112} \left(147+17 \sqrt{105}\right)$$
A: 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*}
