Goal
Show that
$$ \ln (x) \le x - 1 $$
for all $x > 0$.
Proof by inverse functions
Think about their reflection along the line $y = x$ (i.e. their inverse function). Then we get
$$e^x \ge x + 1$$
for all $x \in \Bbb{R}$, which is obviously true.
Proof by definite integrals
Here’s a second proof using definite integrals.
The rectangle $[1,x] \times [0,1]$ contains the region under the graph of the reciprocal $y = 1/x$ in the domain $[1,x]$.
\newcommand\myT{2.75}
\begin{tikzpicture}
\begin{axis}[
axis equal,
axis lines=center,
legend style={at={(1,1)},
anchor=north east,
fill=none},
xlabel={$t$},
ylabel={$y$},
xmin=0,
xmax=4,
xtick=\empty,
ytick=\empty,
extra x ticks={1, \myT},
extra x tick labels={$1$, $x$},
extra y ticks={1},
title={Proof of this inequality with $\log(x) := \int_1^x \frac1t \, \mathrm{d}t$},
]
\addplot[domain=1:\myT, pattern=north east lines, fill opacity=0.3] {1/x}
\closedcycle;
\addlegendentry{$y = \frac{1}{t}$};
\addplot[dashed] coordinates {
(\myT,0)
(\myT,1)
(0,1)
};
\end{axis}
\end{tikzpicture}
The rectangle $[x,1] \times [0,1]$ is contained in the region under the graph of the reciprocal $y = 1/x$ in the domain $[x,1]$. Consider their area and multiply both sides by $-1$ to get the desired inequality.
\newcommand\myT{0.618}
\begin{tikzpicture}
\begin{axis}[
axis equal,
axis lines=center,
legend style={at={(1,1)},anchor=north east,fill=none},
xlabel={$t$},
ylabel={$y$},
xmin=0,
xmax=4,
xtick=\empty,
ytick=\empty,
extra x ticks={\myT,1},
extra x tick labels={$x$,$1$},
extra y ticks={1},
title={Proof of this inequality with $\log(x) := \int_1^x \frac1t \, \mathrm{d}t$}
]
\addplot[domain=\myT:1, pattern=north east lines, fill opacity=0.3]
{1/x} \closedcycle;
\addlegendentry{$y = \frac{1}{t}$};
\addplot[dashed] coordinates {
(1,0)
(1,1)
(0,1)
};
\addplot[dashed] coordinates {
(\myT,0)
(\myT,1)
};
\end{axis}
\end{tikzpicture}