已知集合$T =${ $(x,y)|x > 1,y \in R$}，$M_a =${$(x,y)|y = \left(a\ln x + \frac{1}{a}\right)^2$}；如果存在$(x_0, y_0) \in M_{a_0}$，对于属于$T$且不属于任意$M_a(a \neq 0)$的所有元素$(x,y)$，都有$y – x \leq y_0 – x_0$成立，则$y_0 – x_0$的取值范围是_____.
构造函数释难点,基本不等式+导数定最值
$[8\ln2 – 4, +\infty)$
因为$(x_0,y_0) \in M_{a_0}$，
所以$y_0 = \left(a\ln x_0 + \frac{1}{a}\right)^2$，
$y_0 – x_0 = \left(a\ln x_0 + \frac{1}{a}\right)^2 – x_0 $$\geq \left(2\sqrt{\ln x_0}\right)^2 – x_0 = 4\ln x_0 – x_0$，
设$f(x_0) = 4\ln x_0 – x_0$，
则$f'(x_0) = \frac{4}{x_0} – 1 = \frac{4 – x_0}{x_0}$，
令$f'(x_0) = \frac{4 – x_0}{x_0} > 0 \Rightarrow x_0 \in (1,4)$，$f'(x_0) = \frac{4 – x_0}{x_0} < 0 \Rightarrow x_0 \in (4, +\infty)$，
所以$f(x_0)$在$(1,4)$单调递增，
在$(4, +\infty)$单调递减；
$f(x_0)_{\text{max}} = f(4) = 8\ln2 – 4$，
故$y_0 – x_0 \in [8\ln2 – 4, +\infty)$．
故答案为：$[8\ln2 – 4, +\infty)$
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