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.

$$ x^2+y^2+z^2+xyz=4;\space\space x,y,z>0\qquad (1)\\ (x^2+y)(y^2+z)(z^2+x)+2xyz \le 10\qquad (2)$$
$(1)\Rightarrow 0<x,y,z\le 2 \text{ and}\space x=2\iff y=z=0$
If $x=2$ then $(2)$ is trivially verified.
Put $x=2-\epsilon$ where $0\le \epsilon\le 2$ so $(1)$ becomes
$$y^2+z^2+(2-\epsilon)yz=\epsilon(4-\epsilon)\qquad (1')$$
$(1’)$ is the equation of an ellipse whose axes are contained in the diagonals. In fact, changing coordinates by a rotation of $45^{\circ}$, straightforward calculation in $(1’)$ can transforms $(y,z)$ in $(y_1,z_1)$ giving $$\left(\frac{y_1}{\sqrt{2\epsilon}}\right)^2+\left(\frac{z_1}{2(4-\epsilon)}\right)^2=1$$ 
The concerned points $(y,z)$ are those of the red arc in the figure below corresponding to the value $\epsilon = 0.8$. 

From $(2)$ we get
$$F(\epsilon,y,z)=((2-\epsilon)^2+y)(y^2+z)(z^2+2-\epsilon)+(4-2\epsilon)yz\qquad (2’)$$ where $$\begin{cases}0\le \epsilon\le 2\\ 0<y,z\le \sqrt{\epsilon(4-\epsilon)}\end{cases}$$
For $\epsilon$ fixed, $(2’)$ is maximum when $y=z$ (@HN_NH exercise for) and this occurs when $y^2=\epsilon$ (easily get from $(1’)$ or from the drawn ellipse). It follows
$$F_1(\epsilon)=(2-\epsilon)^3(\epsilon+\sqrt{\epsilon})+(2-\epsilon)^2(\epsilon^2+\epsilon\sqrt{\epsilon})+(2-\epsilon)(3\epsilon+\epsilon\sqrt{\epsilon})+\epsilon^2\sqrt{\epsilon}+\epsilon^2$$
$$F_1(\epsilon)=(2\epsilon^2-6\epsilon+8)\sqrt{\epsilon}+2\epsilon^3-10\epsilon^2+14\epsilon$$
Now, $F_1(\epsilon)$ has in its domain a maximum at $\epsilon=1$ in whose case $F_1(1)=10$. This corresponds to $(x,y,z)=(1,1,1)$ in $(2)$; for the other allowed values the proposed inequality becomes $$(x^2+y)(y^2+z)(z^2+x)+2xyz \lt 10$$
