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.