7. Given a positive real number $k$, for any positive real numbers $a, b$, let $m=\min \left\{a, \frac{b}{k a^{2}+b^{2}}\right\}$, then the maximum value of $m$ is $\qquad$ .
(Note: $\min \{x, y\}$ denotes the smaller of the real numbers $x, y$)
Given $a, \frac{b}{k a^{2}+b^{2}}=\frac{1}{\frac{k a^{2}}{b}+b} \leqslant \frac{1}{2 \sqrt{k} a}$, then the maximum value of $\frac{b}{k a^{2}+b^{2}}$ is $\frac{1}{2 \sqrt{k} a}$, with equality when $b=\sqrt{k} a$. Therefore, when $a=\frac{1}{2 \sqrt{k} a} \Rightarrow a=\frac{1}{\sqrt{2 \sqrt{k}}}=\frac{\sqrt{2}}{2} k^{-\frac{1}{4}}$, the maximum value of $m$ is $\frac{\sqrt{2}}{2} k^{-\frac{1}{4}}$.