How to prove that $(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$, where $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$ $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$, prove
$$(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$$
I try several trig substitutions but feel hopeless with the cyclic term here. The condition $x^2+y^2+z^2+xyz=4$ made it too difficult to homogenize the inequality. I don't even know how to do brutal force either.

$\color{brown}{\textbf{Trigonometrical substitution.}}$
From the given conditions should
\begin{cases}
x,y,z \in [0,2]\\[4pt]
(2z+xy)^2 = (4-x^2)(4-y^2).\tag1
\end{cases}
Taking in account $(1),$ can be applied substitution
$$x=2\sin a,\quad y=2\sin b\quad \Rightarrow \quad z = 2\cos(a+b) = 2\sin c,\tag2$$
where
$$a\ge 0,\quad b\ge 0,\quad c\ge 0,\quad a+b+c=\dfrac\pi2.\tag3$$
Then the given inequality takes the form of
$$4(2\sin^2a+\sin b)(2\sin^2b+\sin c)(2\sin^2c+\sin a) + 8\sin a\sin b\sin c \le 5.\tag4$$
$\color{brown}{\textbf{The proof.}}$
Taking in account $(3),$ one can get
\begin{align}
&2\sin^2 a + \sin b = \sin^2 a + 1 - \cos^2 a + \cos (c+a)\\[4pt]
&= 1 + \cos a (\cos c - \cos a) + \sin a (\sin a -\sin c) 
= 1 - A \sin (a+\varphi),
\end{align}
where
\begin{align}
&A = \sqrt {(\cos c -\cos a)^2 + (\sin a - \sin c)^2} 
= \sqrt {2 - 2\cos (a-c)} = 2 \sin \dfrac{|a-c|}2,\\[4pt]
&\tan \varphi = \dfrac{\cos c -\cos a}{\sin a - \sin c}
= \dfrac{2\sin \frac{a-c}2 \sin \frac{a+c}2}{2\sin \frac{a-c}2 \cos \frac{a+c}2}
= \tan \frac{a+c}2.
\end{align}
Therefore,
$$2\sin^2 a + \sin b = 1 - 2 \sin \frac {|a-c|}2 \sin \frac{3a+c}2.$$
Using the symmetry of task by $a,b,c$, should
$$2\sin^2 a + \sin b \le 1,\quad 2\sin^2 b + \sin c \le 1,\quad
2\sin^2 c + \sin a \le 1.\tag5$$
On the other hand, is known the identity
\begin{align}
\sin(a+b+c) = \cos a \cos b \sin c + \cos a \sin b \cos c + \sin a \cos b \cos c - \sin a \sin b \sin c.
\end{align}
Taking in account $(3),$ one can get
\begin{align}
&\sin a \sin b \sin c = \cos a \cos b \sin c + \cos a \sin b \cos c 
+ \sin a \cos b \cos c - \sin(a+b+c)\\[4pt]
&= \dfrac12(\cos a \sin(b+c) + \cos c \sin(a+b) + \cos b \sin(c+a)) - 1\\[4pt]
&= \dfrac12(\cos^2 a + \cos^2 c + \cos^2 b) - 1
= \dfrac14(\cos 2a + \cos 2b + \cos 2c - 1)\\[4pt]
&= \dfrac18 (\cos(a+b)\cos(a-b) + \cos(b+c)\cos(b-c) + \cos(c+a)\cos(c-a) - 2)\\[4pt]
&= \dfrac18 (\sin c\cos(a-b) + \sin a\cos(b-c) + \sin b\cos(c-a) - 2),
\end{align}
$$\sin a \sin b \sin c \le \dfrac18.\tag6$$
Since from $(5),(6)$ should $(4),$ then the given inequality is proved.
