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.
 A: Since $e^x$ is  convex function (as can be checked with the second derivative), any tangent line will be less than the function.  By taking the tangent at $x=0$, we get $e^x\geq 1+x$.
Now, set $x=b^2-1$.
A: As you suggested, this is through a logarithm. We see that
$$\frac{e^{b^2-1}}{b^2} \geq 1 \leftarrow e^{b^2-1} \geq b^2 \leftarrow \ln(e^{b^2-1}) = b^2-1 \geq \ln(b^2)$$
$$\leftarrow b^2-1 \geq 2\ln(b) \leftarrow b^2 \geq 2\ln(b) + 1.$$
The condition $b^2 \geq 2\ln(b) + 1$ holds for all $b > 0.$
A: 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 $1$.
A: As AccidentalFourierTransform says,
since
$e^x \ge 1+x$,
$e^{b^2-1}
\ge 1+(b^2-1)
= b^2
$.
To show
$e^x \ge 1+x$
for $x \ge 0$,
since
$(e^x)' = e^x$,
integrating from
$0$ to $a$,
$e^a-1
=\int_0^a e^x dx
\ge \int_0^a dx
=a
$
since
$e^x \ge 1$
for $x \ge 0$.
