Complicated Inequality

Reorganising the inequality, we want to find the largest $k$ such that $\Big[(x-y)^2(x+z)(y+z) + (x-z)^2(x+y)(y+z) + (y-z)^2(x+y)(x+z)\Big]^3 \; \ge \; k(x+y)^2(x+z)^2(y+z)^2(x-y)^2(x-z)^2(y-z)^2 \; = \; k(x^2 - y^2)^2(x^2 - z^2)^2(y^2-z^2)^2$ for $x,y,z \ge 0$ with $xy + xz + yz \neq 0$ , and so such that $F(x,y,z) \; = \; \frac{\big[(x-y)^2(x+z)(y+z) + (x-z)^2(x+y)(y+z) + (y-z)^2(x+y)(x+z)\big]^3}{(x^2 - y^2)^2(x^2 - z^2)^2(y^2-z^2)^2} \; \ge \; k$ for $x,y,z \ge 0$ with $xy + xz + yz \neq 0$ and $x,y,z$ distinct. Since $F(x,y,z)$ is a symmetric function of $x,y,z$ , we want to find the largest possible $k$ such that $F(x,y,z) \; \ge \; k \hspace{2cm} 0 \le z < y < x$ (the condition that $xy + xz + yz \neq 0$ simply requires that at least two of $x,y,z$ are nonzero). Since $F(x,y,z)$ is homogeneous of degree $0$ , we may as well assume that $z=1$ , and we want to find the largest possible $k \ge 0$ such that $G(x,y) \ge k$ for $0 \le x < y < 1$ , where $G(x,y) \; = \; F(x,y,1) \;= \; \frac{\big[(x-y)^2(x+1)(y+1) + (x-1)^2(x+y)(y+1) + (y-1)^2(x+y)(x+1)\big]^3}{(x^2-y^2)^2(x^2-1)^2(y^2-1)^2}$ In other words, $k$ is to be the minimum of $G$ over this domain. Looking for turning points of $G(x,y)$ , we find that $\frac{\partial G}{\partial x} \; = \; \frac{8P(x,y)^2}{(x^2 - 1)^3(x^2 - y^2)^3(y^2-1)^2}Q(x,y) \hspace{2cm} \frac{\partial G}{\partial y} \; = \; -\frac{8P(x,y)^2}{(x^2 - 1)^2(x^2 - y^2)^3(y^2 - 1)^3}Q(y,x)$ where $\begin{aligned} \ P(x,y) & = x^3y + xy^3 + x^3 - 2x^2y - 2xy^2 + y^3 - 2xy + x + y \\ & = (1+x+y)(x-y)^2 + (1-xy)\big[x(1-x) + y(1-y)\big] \\ Q(x,y) & = \begin{array}{l} x^6y - 10x^4y^3 + x^2y^5 + x^6 + 4x^5y + 5x^4y^2 - 4x^3y^3 - 2x^2y^4 + 4xy^5 + 5x^4y + 8x^2y^3 + 3y^5 \\ - 10x^4 - 4x^3y + 8x^2y^2 - 4xy^3 - 6y^4 - 2x^2y - 6y^3 + x^2 + 4xy + 3y^2 \end{array} \end{aligned}$ Thus solutions of the equation $\nabla G = \mathbf{0}$ are obtained either by solving $P(x,y) = 0$ or else $Q(x,y) = Q(y,x) = 0$ . It is clear that the only solutions to the equation $P(x,y) = 0$ for $0 \le x,y \le 1$ are $x=y=0$ and $x=y=1$ .

We note that we can write $Q(x,y) + Q(y,x) \; = \; (x+y)R(x,y) \hspace{2cm} Q(x,y) - Q(y,x) \; = \; (x-y)S(x,y)$ for polynomials $R(x,y)$ , $S(x,y)$ .

Suppose that $(\nabla G)(x,y) = \mathbf{0}$ for $0 \le x,y \le 1$ but that $P(x,y) \neq 0$ . This means that $Q(x,y) = Q(y,x) = 0$ , and hence $(x+y)R(x,y) = (x-y)S(x,y) = 0$ . If $x=y$ , this means that either $x=y=0$ or else $R(x,x) = 0$ . But $R(x,x) = -8 (x-1)^3 x (1 + x)^2$ , and hence either $x=y=0$ or $x=y=1$ . These cases can be ignored (since they imply that $P(x,y) = 0$ ). Thus we deduce that $x \neq y$ , and hence that $R(x,y) = S(x,y) = 0$ . The following graph, plotting $R(x,y) = 0$ (blue) and $S(x,y) = 0$ (orange) show that the only points where $R(x,y) = S(x,y) = 0$ simultaneously for $0 \le x,y \le 1$ are $x=y=0$ and $x=y=1$ . In other words, there are no solutions of $(\nabla G)(x,y) = \mathbf{0}$ for $0 \le x,y \le 1$ and $x \neq y$ .

Mathematica backs this up, and tells me that the only solutions to the equations $R(x,y) = S(x,y) = 0$ for $-1 \le x,y \le 1$ are $x=y=0$ and $x=y=1$ . What I have not obtained is an algebraic proof of this, and would welcome any suggestions.

Thus we deduce that there are no turning points for the function $G$ in the domain $0 \le x < y < 1$ . Since $G(x,y) \to \infty$ either as $y \to 1$ or as $y-x \to 0$ , we deduce that the minimum for $G$ on this domain must be achieved on the boundary $x=0$ . Thus, to minimize $G(x,y)$ for $0 < x < y \le 1$ , we need simply to minimize $G(0,y) \; = \; \frac{8 (y^2 + 1)^3}{y (y^2-1)^2} \hspace{2cm} 0 < y \le 1$ The minimum is achieved at $y = \sqrt{3} - \sqrt{2}$ , and $G\big(0,\sqrt{3}-\sqrt{2}\big) = 24\sqrt{3}$ .

Thus we deduce that $k = 24\sqrt{3}$ .

I suspect that it is possible, for any $0 \le x,y \le 1$ , to find $0 \le v \le 1$ such that $G(x,y) \ge G(0,v)$ can be proved easily. If $v$ does exist, and an explicit form for this $v$ can be found, the whole business would be much simpler!