Consider the one-dimensional SDE $$ dX_t = f(X_t)dt + \sigma(X_t)dW_t,\quad X_0 = 1, $$ where $W_t$ is a Brownian motion under the measure $\mathbb{P}$, in its natural filtration. Suppose that $f(x)-\sigma^2(x)/2x=0$ and $\sigma^2(x)\le x^2$. Show that $X_t>0$ almost surely.
My initial attempt was to consider the log-process $\log(X_t)$. Using Ito's lemma one finds $$ d\log(X_t) = \frac{1}{X_t}\left(f(X_t) - \frac{\sigma^2(X_t)}{2X_t}\right)dt + \frac{\sigma(X_t)}{X_t}dW_t = \frac{\sigma(X_t)}{X_t}dW_t. $$ Integrating now leads to $$ \log(X_t)=\int_0^t\frac{\sigma(X_s)}{X_s}dW_s. $$ By the Ito isometry, the variance of the above process is bounded by $t$: $$ \text{Var}\big[\log(X_t)\big] = \mathbb{E}\left[\int_0^t\frac{\sigma^2(X_s)}{X_s^2}ds\right] \le t. $$ I feel like this is heading in the right direction, but I'm struggling to see what the next step is. Any suggestions would be very welcome.