2. 已知正数 $a$，$b$ 满足 $2a^4+\dfrac{2}{b^4}\le 2(\ln 2a-\ln b)+1$，则 $a^2+b^2=$__.
解
$2a^4+\dfrac{2}{b^4}\le 2(\ln 2a-\ln b)+1$
$\therefore 2a^4+\dfrac{2}{b^4}\le \ln\dfrac{4a^2}{b^2}+1$
$\therefore \ln\dfrac{4a^2}{b^2}+1\ge 2\sqrt{2a^4\cdot\dfrac{2}{b^4}}=\dfrac{4a^2}{b^2}$（当且仅当 $ab=1$ 时等号成立）
设 $f(x)=\ln x-x+1$，则 $f'(x)=\dfrac{1}{x}-1=\dfrac{1-x}{x}$
$\therefore 0<x<1$ 时 $f(x)$ 单增，$x>1$ 时 $f(x)$ 单减
$\therefore f(x)_{\text{极大值}}=f(1)=0$
$\therefore f\left(\dfrac{4a^2}{b^2}\right)\le 0$
$\therefore \ln\dfrac{4a^2}{b^2}+1\le \dfrac{4a^2}{b^2}$
$\therefore \ln\dfrac{4a^2}{b^2}+1=\dfrac{4a^2}{b^2}$
$\therefore \dfrac{4a^2}{b^2}=1$
$\therefore \begin{cases} ab=1 \\ 2a=b \end{cases}$，$\therefore \begin{cases} a=\dfrac{\sqrt{2}}{2} \\ b=\sqrt{2} \end{cases}$
$\therefore a^2+b^2=\dfrac{5}{2}$