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.

Taking the derivative of $\Phi(x)$,
$$
\Phi'(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\tag{1}
$$
For $x\ge0$, everything is non-negative, so $x\Phi(x)+\Phi'(x)\ge0$.
For $x<0$,
$$
\begin{align}
\Phi(x)
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-t^2/2}\;\mathrm{d}t\\
&\le\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-t^2/2}\frac{t}{x}\;\mathrm{d}t\\
&=-\frac{1}{x}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\\
&=-\frac{1}{x}\Phi'(x)\tag{2}
\end{align}
$$
Therefore, we also get $x\Phi(x)+\Phi'(x)\ge0$.
