Prove $(x+y)(y^2+z^2)(z^3+x^3) < \frac92$ for $x+y+z=2$ $x,y,z \geqslant 0$ and $x+y+z=2$, Prove
$$(x+y)(y^2+z^2)(z^3+x^3) < \frac92$$
While numerical method can solve this problem, I am more interested in classical solutions. I tried this problem for the past few months, using all kinds of AM-GM and CS, but still cannot solve it. I hope someone can help me out with this one.
 A: Let $f(x,y,z)=(x+y)(y^2+z^2)(z^3+x^3)$.
Assume that $y$ is largest of $x,y,z$. Then $$f(x,y,z)-f(x,z,y)=(y^2+z^2)(x + y) (x + z) (y - z) (x - y - z)\le0$$and therefore we may assume $x$ or $z$ is largest of $x, y, z$.
If $x$ is largest, then
$$f(x,y,z)-f(z,y,x)=(z^3+x^3)(z-x)(xy+xz-y^2+yz)\le0$$and therefore we may assume that $z$ is largest of $x,y,z$.
Now,$$f(0,x+y,z)-f(x,y,z)=(x+y)(((x+y)^2+z^2)z^3-(y^2+z^2)(z^3+x^3))\\=x (x+y)(-x^2 y^2 - x^2 z^2 + x z^3 + 2 y z^3)\ge0$$and we may assume that $x=0$.
Now it is just $y(y^2+(2-y)^2)(2-y)^3<\frac{9}{2}$, or$$2 y^6 - 16 y^5 + 52 y^4 - 88 y^3 + 80 y^2 - 32 y + 4.5>0$$which is true for all nonnegative $y$.
A: The proof comes by considering three cases.
Case 1: $z>1$, $y> 0.1$. We have
$$(x+y)(y^2+z^2)(z^3+x^3) - \frac92= (2-z)((2-x-z)^2+z^2)(z^3+(2-z-y)^3) - \frac92\\
< (2-z)((2-z)^2+z^2)(z^3+(2-z-0.1)^3) - \frac92$$
which is less than 0, see this plot:

and in more detail here:

Case 2: $z>1$, $y\leq 0.1$. We have
$$(x+y)(y^2+z^2)(z^3+x^3) - \frac92= (2-z)(y^2+z^2)(z^3+(2-z-y)^3) - \frac92\\
< (2-z)((0.1)^2+z^2)(z^3+(2-z)^3) - \frac92$$
which is less than 0, see this plot:

Case 3:
$z \leq 1$. Here, let $x = \frac{r}{2} (1+q)$ and $y = \frac{r}{2} (1-q)$. Hence, $x+y = r$  and $1 \leq r \leq 2$. Further, $-1\leq q \leq 1$ since $x \geq 0$ and $y \geq 0$. With this notation, we have
$$(x+y)(y^2+z^2)(z^3+x^3) - \frac92= r (\frac{r^2}{4} (1-q)^2+(2-r)^2)((2-r)^3+\frac{r^3}{8} (1+q)^3) - \frac92$$
which is less than 0 in the  given ($q,r$)-range, see this contourplot:

Note 1: since no upper limits have been made in rewriting the equation for case 3, it would have been possible to drop the condition $z\leq 1$ and, giving the proof without case discussion, to draw the contourplot for the full range $0\leq r \leq 2$. However, the delicate case where the bound is very tight occurs in case 1 and is easier to see with the given description in case 1. In a contourplot for the full range $0\leq r \leq 2$, this delicate case can only be shown with magnification of the relevant area.
Note 2: The first two cases are in one parameter $z$ only, so showing the negativity over the given $z$-range will be easily possible, instead of plots, by analytic approximations.
Note 3: In the third case, the values are manifestly negative in the whole considered range. Hence, also here, showing negativity over the whole range will be easily possible, instead of plots, by coarse approximations.    
This completes the proof.
