Proof that $x \Phi(x) + \Phi'(x) \geq 0$ $\forall x$, where $\Phi$ is the normal CDF As title. Can anyone supply a simple proof that
$$x \Phi(x) + \Phi'(x) \geq 0 \quad \forall x\in\mathbb{R}$$
where $\Phi$ is the standard normal CDF, i.e.
$$\Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-y^2/2} {\rm d} y$$

I have so far:
Defining $f(x) = x \Phi(x) + \Phi'(x)$ we get
$$
\begin{align}
f'(x) & = \Phi(x) + x \Phi'(x) + \Phi''(x) \\
& = \Phi(x) + x\Phi'(x) - x\Phi'(x) \\
& = \Phi(x) \\
& >0
\end{align}$$
so it seems that if we can show
$$\lim_{x\to-\infty} f(x) = 0$$
then we have our proof - am I correct?
Clearly $f$ is the sum of two terms which tend to zero, so maybe I have all the machinery I require, and I just need to connect the parts in the right way! Assistance will be gratefully received.

In case anyone is interested in where this question comes from:
Bachelier's formula for an option struck at $K$ with time $T$ until maturity, with volatility $\sigma>0$ and current asset price $S$ is given by
$$V(S) = (S - K) \Phi\left( \frac{S-K}{\sigma S \sqrt{T}} \right) + \sigma S \sqrt{T} \Phi' \left( \frac{S-K}{\sigma S \sqrt{T}} \right) $$
Working in time units where $\sigma S\sqrt{T} = 1$ and letting $x=S-K$, we have
$$V(x) = x \Phi(x) + \Phi'(x)$$
and I wanted a simple proof that $V(x)>0$ $\forall x$, i.e. an option always has positive value under Bachelier's model.

I will concentrate at $\lim_{x \to -\infty} f(x) = 0$, where $f(x) = x \Phi(x) + \Phi'(x)$.
Consider
$$
  g(x) = \sqrt{2 \pi}  f(-x) = \mathrm{e}^{-\frac{x^2}{2}} - x \int_{x}^\infty \mathrm{e}^{-\frac{y^2}{2}} \mathrm{d} y  \qquad \text{for} \qquad x > 0.
$$
Then
$$
   g(x) = \mathrm{e}^{-\frac{x^2}{2}} - x \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \frac{\mathrm{d} t}{\sqrt{2t}} \, > \, \mathrm{e}^{-\frac{x^2}{2}} -  \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \mathrm{d} t = 0
$$
where $ x \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \frac{\mathrm{d} t}{\sqrt{2t}} < \frac{x}{\sqrt{2 \frac{x^2}{2}}} \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \mathrm{d} t = \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \mathrm{d} t = \mathrm{e}^{-\frac{x^2}{2}}$ for $x>0$.
On the other hand $g(x) < \mathrm{e}^{-\frac{x^2}{2}}$ by definition, so $\lim_{x \to \infty} g(x) = \lim_{x \to \infty} \sqrt{2 \pi} f(-x)$ vanishes being sandwiched between 0 and $\mathrm{e}^{-\frac{x^2}{2}}$ which also tends to zero for large $x$.
