How to prove that $\frac {e^{b^2-1}}{b^2}$ ≥ 1

Question

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.

Answer 1

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 $x=b^2-1$.

Answer 2

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 $b > 0.$

Answer 3

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 $1$.

Answer 4

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 $x \ge 0$.