How to prove that $\frac {e^{b^2-1}}{b^2}$ ≥ 1 How to prove that $$\frac {e^{b^2-1}}{b^2} \ge 1?$$ Use logarithm or limit or what? Or do we have to use it as a conclusion to prove it backwards? And how to prove it forwards, that is, without assuming this is right.

First, let's make a substitution: $x=b^2$
The expression now becomes $\displaystyle\frac{e^{x-1}}{x}$.
Next, let's take the derivative of this expression:
$\displaystyle \frac{d}{db}\frac{e^{x-1}}{x}=\frac{(x-1)e^{x-1}}{{x^2}}$
We know that local maxima/minima of this expression occur at values of $x$ for which $\displaystyle\frac{(x-1)e^{x-1}}{{x^2}}=0$
We also know that $x$ must be non-negative, so we only have to consider non-negative solutions to the above equation. The only positive solution is $x=1$. 
Since the derivative $\displaystyle\frac{(x-1)e^{x-1}}{{x^2}}$ is positive for $1<x$ and negative for $0\leq x<1$, a minimum (as opposed to a maximum) must occur at $x=1$. 
Ergo, the smallest value of $\displaystyle\frac{e^{b^2-1}}{b^2}$ occurs when $b^2=1$, and the smallest value of $\displaystyle\frac{e^{b^2-1}}{{b^2}}$ is $1$.
