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


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?
