@MarcPalm Yes, but I maintained the notation in Guinand

@Hans can you send me a mail. To send you this privately. They want to pass the discussion to chat. I prefer mail. My address is easy to get.

For $x>0$ in the case of discriminant positive, the inequality will be true if $x>$ the greater of the two roots. I think this is also promising.

@Did yes I was wrong. This proves the inequality for $|x|<0.597. And the rest is not so easy, because now the intervals are infinite. I changed my solution correspondingly.

@Hans I had edited my answer because at the start I said $|x|<0.597$ instead of the intended $|x|>0.597$.

@Hans It is realy easy. But now I am a little confused about who is interested.

but your function has a term $-(x+N)^2$ instead of $-(x N+n)^2$.

@user64494 My answer proof completely the inequality for $x<-0.597$ or $x>0.597$. For this you have no need of a bound of the derivative. Now to show $f(x)>0$ (my $f$ is different from yours) on the interval $|x|<0.597$ you may apply the maximal slope principle. This need a bound of the derivative on $|x|<0.597$ (a rough bound suffice). This is very easy to get. And you finish without difficulty the proof with a little computation (see the paper cited in my answer).

@user64494 To apply the maximal slope principle you only need a rough bound of the derivative. For example substitute all exp(-x^2/2) by 1 and all N(x) by 1, all x by 0.597. All in absolute value and this bound will suffice.