Proof that $t-1-\log t \geq 0$ for $t > 0$ Using basic calculus, I can prove that $f(t)=t-1-\log t \geq 0$ for $t > 0$ by setting the first derivative to zero
\begin{align}
\frac{df}{dt} = 1 - 1/t = 0
\end{align}
And so I have a critical point at $t=1$ and $f(1)=0$. Then I calculate the second derivative $\frac{d^2f}{dt^2} = 1/t^2 \geq 0$ meaning that $f$ is a convex function with a minimum value of 0 so $f \geq 0$ for $t > 0$.
However, something in my gut tells me there's a way to prove this without even using the first or any derivative of $f$. I've been thinking about this for a while and I haven't been able to do this.
Question is: can you prove $f\geq 0$ without relying on any derivatives of $f$?
 A: With the definition of $\log$ as an integral
$$t-1 -\log t = \int_1^t \frac{x-1}{x} dx.$$
Because the integrand is positive for $x>1$, we have $t-1 -\log t \geq 0$ for $t>1$.
Because the integrand is negative for $x<1$, we have $t-1 -\log t \geq 0$ for $0<t<1$.
A: *

*With the definition of $\log t$ as an integral. We can define 
$$\log t = \int_1^t \frac{1}{x}\,dx.$$
The function $\frac{1}{x}$ is decreasing, so a left hand sum approximation is always an over estimate, and a right hand sum approximation is always an underestimate. Dividing the interval $[1,t]$ into a single interval of length $t-1$ and evaluating at the left endpoint, we get
$$\ln(t) = \int_1^t\frac{1}{x}dx \leq f(1)(t-1) = t-1$$
giving the desired inequality. 
If $0\lt t\lt 1$, then we first switch limits and use a right hand sum with one interval we get:
$$\ln(t) = \int_1^t\frac{1}{x}dx = -\int_t^1\frac{1}{x}dx \leq -f(1)(1-t) = t-1$$
(we have $\int_t^1\frac{1}{x}dx \geq f(1)(1-t)$ since the right hand sum is an underestimate), so multiplying by $-1$ gives the inequality above), giving the desired inequality again.
If $t=1$, the inequality reduces to $0\geq \log(1)$, which is of course true. 

*With exponentials. $t-1-\log t\geq 0$ if and only if $\log t\leq t-1$, if and only if $t \leq e^{t-1}$.


*

*With the Taylor series definition of $e^t$. Since
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$
then
$$e^{t-1} = 1 + (t-1) + \frac{(t-1)^2}{2!} + \frac{(t-1)^3}{3!} + \cdots.$$
If $t\geq 1$, then $e^{t-1}\geq 1+(t-1) = t$, giving the desired inequality. If $0\lt t\lt 1$, then we have an alternating series
$$ \frac{(t-1)^2}{2!} + \frac{(t-1)^3}{3!} + \frac{(t-1)^4}{4!}+\cdots$$
with ever decreasing terms:
$$\frac{|t-1|^{n+1}}{(n+1)!} \lt \frac{|t-1|^n}{n!} \Longleftrightarrow |t-1|\lt n+1,$$
which holds because $n\geq 2$ and $|t-1|\lt 1$. Thus, the "tail" (starting in the quadratic term) of the series is positive, so $e^{t-1} \geq 1+(t-1) = t$ still holds, giving the desired inequality as well.

*With the definition of $e^t$ as a limit. We have
$$e^x = \lim_{n\to\infty}\left(1 + \frac{x}{n}\right)^n$$
so
$$e^{t-1} = \lim_{n\to\infty}\left(1 + \frac{t-1}{n}\right)^n.$$
If $t\geq 1$, the sequence is nondecreasing (we are compounding interest, so the more often we compound the bigger the payoff). In particular, $e^{t-1}\geq 1 +\frac{t-1}{1} = t$, giving the desired inequality. If $0\lt t \lt 1$, then we have 
$$e^{t-1} = \lim_{n\to\infty}\left(1 - \frac{1-t}{n}\right)^n.$$
Again, the sequence is increasing (this is like paying off a debt with fixed interest; if you pay down the capital more often, your total interest will be smaller in the end). So again we have $e^{t-1} \geq 1 - \frac{1-t}{1} = t$.
