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.

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 $x \ge 0$.
