If the inequality $x^{2}-\log _{m} x<0$ holds for all $x$ in $\left(0, \frac{1}{2}\right)$, find the range of real values for $m$.

---

The inequality $x^{2}-\log _{m} x<0$ must hold for all $x$ in the interval $\left(0, \frac{1}{2}\right)$. Determine the range of real values for $m$.
Consider the function $f(x)=x^{2}-\log _{m} x, x \in\left(0, \frac{1}{2}\right]$, then $f(x)1$, then $x^{2}-\log _{n} x>x^{2}-\log _{m} 1=x^{2}>0$, a contradiction. Therefore, $0<m<1$. By the monotonicity of $x^{2}-\log _{n} x$, we know that $f(x)$ is increasing, so $f(x)<0$ for all $x \in\left(0, \frac{1}{2}\right)$ if and only if $f\left(\frac{1}{2}\right) \leqslant 0$. From $f\left(\frac{1}{2}\right) \leqslant 0$, we get $\left(\frac{1}{2}\right)^{2}-\log _{m} \frac{1}{2} \leqslant 0, \log _{m} m^{\frac{1}{4}} \leqslant \log _{m} \frac{1}{2}$. Since $0<m<1$, we have $m^{\frac{1}{4}} \geqslant \frac{1}{2}$, which means $\frac{1}{16} \leqslant m<1$. Therefore, the range of $m$ is $\left[\frac{1}{16}, 1\right)$.