Inequality with a rational function

Question

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:

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

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

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

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

Answer 1

For convenience, to simplify, take the sixth root of $f$ to give $$g(x,y,z)=\frac{(a_1(x+y+z)^3+a_2x^3+a_3(x+y)^3+a_4(y+z)^3+a_5z^3)^{1/3}}{(a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2)^{1/2}}$$ which is the ratio of a weighted 3-norm to its 2-norm. Notice also that the function is homogeneous in $x,y,z$, so one can divide the numerator and denominator by any scalar, $$g(x,y,z)=\frac{(a_1+a_2X^3+a_3(X+Y)^3+a_4(1-X)^3+a_5(1-X-Y)^3)^{1/3}}{(a_1+a_2X^2+a_3(X+Y)^2+a_4(1-X)^2+a_5(1-X-Y)^2)^{1/2}}$$ where $X=x/(x+y+z)$, $Y=y/(x+y+z)$.

A complete solution will take up too much space but here's the main idea. Suppose you want to maximize $(\alpha U^3+\beta V^3)^{1/3}/(\alpha U^2+\beta V^2)^{1/2}$, with $\alpha<\beta$, then that is the same as maximizing $$\alpha U^3+\beta V^3\qquad \mathrm{such\ that} \qquad \alpha U^2+\beta V^2=1.$$ If you use Lagrange multipliers, we find $$3\alpha U^2=2\lambda\alpha U,\qquad 3\beta V^2=2\lambda\beta V$$ so either $U=0,V=\beta^{-1/2}$, or $U=\alpha^{-1/2},V=0$, or $U=V=(\alpha+\beta)^{-1/2}$. The function evaluated at each of these points is equal to $\beta^{-1/6}$, $\alpha^{-1/6}$, $(\alpha+\beta)^{-1/6}$ respectively. The maximum is the middle one, with $U\ne0,V=0$. In general, the components associated with the largest coefficients need to be made $0$.

Back to the main problem, we need to make $1-X-Y=0$, so $X+Y=1$; then also $1-X=0$, so $X=1$. Hence $X=1$, $Y=0$, which is the same as $y=0$, $x=x+y+z$, so $z=0$. Any vector of the form $(x,0,0)$ will achieve the maximum value.

Answer 2

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 $$\frac{(a_1(x+y+z)^3 + a_2x^3 + a_3(x+y)^3 + a_4(y+z)^3 + a_5z^3)^2}{(a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2)^3} \le \frac{1}{a_1+a_2+a_3}.$$