Prove that $\sum\limits_{\mathrm{cyc}} xy^2+xyz \Big [1+\frac23 \sum\limits_{\mathrm{cyc}} (x^2 - xy) \Big] \leq 4$ 
Let $x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$. Prove that
$$xy^2+yz^2+zx^2+xyz \Big [1+\frac23 (x^2+y^2+z^2-xy-yz-zx) \Big] \leq 4  .$$

Die a lot of brain cell with this one ( 4 years already). I don't think the usual trig or algebraic substitutes work for this inequality !!!
 A: Let $f=xy^2+\cdots+xyz(\cdots),g=x^2+y^2+z^2+xyz-4$
If we use an optimization software, then we get $\max(f)$ when $g=0,x,y,z\geq 0$, that is $4$ only for $x=y=z=1$.
Possibly the software (because of the gradient methods) may forget values of $(x,y,z)$ improving the maximum. If we want to be sure of the result, we can use the Lagrange method; but often the calculations are very complicated (here we consider 4 algebraic equations with 4 unknowns). I used the computer to solve the system.
$\bullet$Case 1.$(x,y,z)$ is not in the edge of $(\mathbb{R}^+)^3$, that is $x,y,z>0$
Then the $(4\times 4)$ system is $\dfrac{\partial f}{\partial x}+a\dfrac{\partial g}{\partial x}=0,\cdots,g=0$ with unknows $x,y,z,a$.
This system has $46$ solutions in $\mathbb{C}^3$, but only one of them is $>0: (1,1,1)$
$\bullet$Case 2.For example $x=0$, Then we search $\max(yz^2)$ when $y^2+z^2=4,y,z\geq 0$.
We substitute $y=2\cos(u),z=2\sin(u)$ where $u\in [0,\pi/2]$ and it is easy to see that the maximum of $8\cos(u)\sin(u)^2$ is $\dfrac{16}{3\sqrt{3}}<4$.
A: Too long for a comment :
I use the substitution @Andreas .
I try to eliminate a part of the inequality and it seems we have the inequalities for $u,v,w\in(0,1]$ such that  :
$$a=2\sqrt{\frac{uv}{\left(w+u\right)\left(w+v\right)}},b=2\sqrt{\frac{uw}{\left(v+u\right)\left(w+v\right)}},c=2\sqrt{\frac{wv}{\left(w+u\right)\left(u+v\right)}}$$ then it seems we have
$$g(a,b,c)\leq f(a,b,c)\leq 4$$
Where :
$$g(a,b,c)=ab^{2}+bc^{2}+ca^{2}+abc\left(1+\frac{2}{3}(a^{2}+b^{2}+c^{2}-ab-bc-ca)\right)$$
And :
$$f(a,b,c)=\left(1-abc\right)\left(ab^{2}+bc^{2}+ca^{2}\right)+\left(2abc-1\right)+abc+2$$
I don't know if it's simpler but let's try it .
Hope it helps in some way .
