Show that this stochastic process is a.s. strictly positive. 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. 

Define stopping times by
$$T_k := \inf\{t \geq 0; X_t \leq k^{-1}\}, \qquad k \in \mathbb{N}$$
and 
$$T := \inf\{t \geq 0; X_t \leq 0\}.$$
Following the reasoning in your question we find that
$$\mathbb{E} \left( \left| \log(X_{t \wedge T_k}) \right|^2 \right) \leq t.$$
By the continuity of the sample paths of $(X_t)_{t \geq 0}$, we have $X_{T_k}(\omega) = k^{-1}$ for $\omega \in \{T_k<\infty\}$, and therefore we get that
$$\mathbb{E} \bigg( 1_{\{T_k \leq t\}} \underbrace{|\log(X_{T_k})|^2}_{=|\log(k^{-1})|^2} \bigg) \leq t.$$
Since $\{T \leq t\} \subseteq \{T_k \leq t\}$ this shows, in particular, that
$$\mathbb{E}\left(1_{\{T \leq t\}} |\log(k^{-1})|^2 \right) \leq t,$$
i.e.
$$\mathbb{P}(T \leq t) \leq \frac{t}{|\log(k^{-1})|^2}.$$
As $|\log(k^{-1})|^2 \to \infty$ as $k \to \infty$ we conclude that $\mathbb{P}(T \leq t)=0$. As $t>0$ is arbritrary, this proves the assertion.
