3．设实数$a \neq 0$，不等式$\frac{\text{e}^{x}}{a} - 2ax \geq \sqrt{2\text{e}^{x} + 1}$对任意实数$x \geq - \frac{1}{2}$恒成立，则a的取值范围为\\\\\\\\\\．
解：令$x = 0$，得$0 < a \leq \frac{\sqrt{3}}{3}$．
下证：对任意的$a \in \left( 0,\frac{\sqrt{3}}{3} \right\rbrack$，不等式$\frac{\text{e}^{x}}{a} - 2ax \geq \sqrt{2\text{e}^{x} + 1}$恒成立．
令$g(a) = \frac{\text{e}^{x}}{a} - 2ax - \sqrt{2\text{e}^{x} + 1}$
①当$x > 0$时，$g(a)$单调递减，所以$g(a) \geq g\left( \frac{\sqrt{3}}{3} \right) = \sqrt{3}\text{e}^{x} - \frac{2\sqrt{3}}{3}x - \sqrt{2\text{e}^{x} + 1}$
令$t = \text{e}^{x}$，则$x = \ln t$，则只需证明$\varphi(t) = \sqrt{3}t - \frac{2\sqrt{3}\ln t}{3} - \sqrt{2t + 1} \geq 0$在$(0, + \infty)$上恒成立
由$\varphi'(t) = \sqrt{3} - \frac{2\sqrt{3}}{3t} - \frac{1}{\sqrt{2t + 1}}$，可知$\varphi'(t)$单调递增，且$\varphi'(1) = 0$，故$\varphi(t)$在$(0,1)$上单调递减，在$(1, + \infty)$上单调递增，
所以$\varphi(t) \geq \varphi(1) = 0$成立；
②当$- \frac{1}{2} \leq x < 0$时，$- \frac{\text{e}^{x}}{x} \geq \frac{\text{e}^{- \frac{1}{2}}}{\frac{1}{2}} = \frac{2}{\sqrt{\text{e}}} > 2a^{2}$，$g(a)$单调递减，$g(a) \geq g\left( \frac{\sqrt{3}}{3} \right) = \sqrt{3}\text{e}^{x} - \frac{2\sqrt{3}}{3}x - \sqrt{2\text{e}^{x} + 1}$
由①可知$\varphi(t)$在$\left\lbrack - \frac{1}{2},0 \right)$上单调递减，所以$\varphi(t) > \varphi(0) > \varphi(1) = 0$成立．
综上，得证$a \in \left( 0,\frac{\sqrt{3}}{3} \right\rbrack$．
故答案为：$\left( 0,\frac{\sqrt{3}}{3} \right\rbrack$.