Inequality with a rational function Let $0< a_1\leq a_2\leq a_3\leq a_4\leq a_5$. Define the function $$f(x,y,z)=\cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_4(y+z)^3+a_5(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_4(y+z)^2+a_5(z)^2 )^3}.$$
Show that for all $x,y,z\in \mathbb R$ we have $f(x,y,z)\leq \cfrac{1}{a_1+a_2+a_3}.$
Information about the problem:


*

*The function forms an asymptote at $\cfrac{1}{a_1+a_2+a_3}$.

*We have $f(1,0,0)=\cfrac{1}{a_1+a_2+a_3}$.

*As a function of $x$, it seems there are only three critical points. Same for $y$ and $z$.

*Perhaps it helps to show that
$$\cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_4(y+z)^3+a_5(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_4(y+z)^2+a_5(z)^2 )^3}\leq \cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_3(y+z)^3+a_3(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_3(y+z)^2+a_3(z)^2 )^3}.$$

Not an answer. 
First, I would like to give the equivalent problem:
Problem 1: Let $0 < a_1 \le a_2 \le a_3 \le a_4 \le a_5$. Let $x, y, z \in \mathbb{R}$ such that
$$a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2 = a_1 + a_2 + a_3.\tag{1}$$ Prove that
$$a_1(x+y+z)^3 + a_2x^3 + a_3(x+y)^3 + a_4(y+z)^3 + a_5z^3 \le a_1 + a_2 + a_3.$$
Remark: The constraint (1) is an ellipsoid. 
Second, if $x, y, z$ is non-negative, the inequality is true as Lemma 1 stated. One may use the Buffalo Way to prove Lemma 1 (easily with computer). So, if one can prove that the optimal $x, y, z$ in the original problem are non-negative, the desired result follows. 
Lemma 1: Let $x, y, z\ge 0$ not all zero. Then
$$\frac{(a_1(x+y+z)^3 + a_2x^3 + a_3(x+y)^3 + a_4(y+z)^3 + a_5z^3)^2}{(a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2)^3}
\le \frac{1}{a_1+a_2+a_3}.$$
