China Mathematical Competition

The given condition is equivalent to $$(a{x}^2{y}^2+b)+(a(x+y{)}^2+b)\ge (a{x}^2+b)(a{y}^2+b).①$$ In ①, let $y=0$. We have $b+(a{x}^2+b)\ge (a{x}^2+b)\cdot b$, or $$(1-b)a{x}^2+b(2-b)\ge 0.$$ As $a>0$ and $a{x}^2$ can be sufficiently large, then $1-b\ge 0$, i.e., $0<b\le 1$. In ①, let $y=-x$. We have $(a{x}^4+b)+b\ge (a{x}^2+b{)}^2$, or $$(a-a^2)x^4-2ab{x}^2+(2b-b^2)\ge 0.②$$ Denote the left-hand side of ② as $g(x)$. It is obvious that $a-a^2\ne 0$ (otherwise, from $a>0$ we know $a=1$. Then $g(x)=-2b{x}^2+(2b-b^2)$ with $b>0$, which means $g(x)$ can be negative. A contradiction). Then $$\begin{array}{rl}g(x)& =(a-a^2){\left(x^2-\frac{ab}{a-a^2}\right)}^2-\frac{(ab{)}^2}{a-a^2}+(2b-b^2)\\ & =(a-a^2){\left(x^2-\frac{b}{1-a}\right)}^2+\frac{b}{1-a}(2-2a-b)\\ & \ge 0\end{array}$$ holds for any real number $x$. So we have $a-a^2>0$, i.e., $0<a<1$. Furthermore, from $\frac{b}{1-a}>0$ and $$g\left(\sqrt{\frac{b}{1-a}}\right)=\frac{b}{1-a}(2-2a-b)\ge 0,$$ we have $2a+b\le 2$. So far, we get the necessary condition that $a,b$ must satisfy as follows: $$0<b\le 1,0<a<1,2a+b\le 2.\stackrel{◯}{3}$$ We are going to prove that for any pair $(a,b)$ satisfying ③ and any real numbers $x,y$, ① holds, or equivalently, $$h(x,y)=(a-a^2)x^2{y}^2+a(1-b)(x^2+y^2)+2axy+(2b-b^2)\ge 0.$$ As a matter of fact, when ③ holds, we then have $$a(1-b)\ge 0,a-a^2>0\text{and}\frac{b}{1-a}(2-2a-b)\ge 0.$$ Combining it with $x^2+y^2\ge -2xy$, we get $$\begin{array}{rl}h(x,y)& \ge (a-a^2)x^2{y}^2+a(1-b)(-2xy)+2axy+(2b-b^2)\\ & =(a-a^2)x^2{y}^2+2abxy+(2b-b^2)\\ & =(a-a^2){\left(xy+\frac{b}{1-a}\right)}^2+\frac{b}{1-a}(2-2a-b)\ge 0.\end{array}$$ Therefore, the set of all the pairs $(a,b)$ meeting the given condition is $$\{(a,b)\mid 0<b\le 1,0<a<1,2a+b\le 2\}.$$