$m x e^{m x^2}$ and $\ln x$

Unlike the problem question suggest there is only one value of $m$ , let it be $m_1$ , that satisfies $mxe^{mx^2} \ge \ln x$ for all real $x > 0$ .

Let us consider the equation $mxe^{mx^2} = \ln x$ . We find that (see graphs) for $m > m_1$ , there is no solution to the equation or $mxe^{mx^2} > \ln x, \forall x > 0$ . When $m=m_1$ , there is only one solution and $mxe^{mx^2} \ge \ln x, \forall x > 0$ is true. For $m < m_1$ the inequality does not hold.

When $m=m_1$ , we have $m_1xe^{m_1x^2} = \ln x\ ...(*)$ and

$\begin{aligned} \frac d{dx} m_1xe^{m_1x^2} & = \frac d{dx} \ln x \\ m_1e^{m_1x^2} + 2m_1^2x^2e^{m_1x^2} & = \frac 1x & \small \blue{\text{As }(*): m_1xe^{m_1x^2} = \ln x} \\ \ln x + 3m_1x^2 \ln x & = 1 \\ \implies m_1 & = \frac {1-\ln x}{2x^2\ln x} \\ \implies (*): \quad \frac {1-\ln x}{2x\ln x} e^{\frac {1-\ln x}{2\ln x}} & = \ln x \\ (1-\ln x)e^{\frac {1-\ln x}{2\ln x}} & = 2x \ln^2 x \\ \implies x & = \sqrt e & \small \blue{\text{I solved it using numerical method.}} \\ \implies m_1 & = \boxed{\frac 1{2e}} \end{aligned}$