The inequality \(\dfrac 1 t \le 1\) holds for \(t\ge1\).
By integrating both sides of this inequality
over the interval \(1\le t \le x\), show that
\[
\ln x \le x-1
\tag{\(*\)}
\]
for \(x \ge 1\). Show similarly
that \((*)\) also holds for \(0 < x \le 1\).
Starting from the inequality
\(\dfrac{1}{t^2} \le \dfrac1 t \) for \(t \ge 1\),
show that
\[
\ln x \ge 1-\frac{1}{x}
\tag{\(**\)}
\]
for \(x > 0\).
Show, by integrating (\(*\)) and (\(**\)), that
\[
\frac{2}{ y+1} \le \frac{\ln y}{ y-1} \le \frac{ y+1}{2 y}
\]
for \( y > 0\) and \( y\ne1\).
Solution:
\(\,\)
\begin{align*}
(x \geq 1): && \int_1^x \frac{1}{t} \d t &\leq \int_1^x 1 \d t \\
\Rightarrow && \ln x - \ln 1 &\leq x - 1 \\
\Rightarrow && \ln x & \leq x - 1 \\
\\
(0 < x \leq 1):&& \int_x^1 1\d t &\leq \int_x^1 \frac{1}{t} \d t \\
\Rightarrow&& 1- x &\leq \ln 1 - \ln x \\
\Rightarrow&& \ln x &\leq x - 1
\end{align*}
\(\,\)
\begin{align*}
(x \geq 1): && \int_1^x \frac{1}{t^2} \d t &\leq \int_1^x \frac{1}{t} \d t \\
\Rightarrow && -\frac1x+1 &\leq \ln x - \ln 1 \\
\Rightarrow && 1 - \frac1x &\leq \ln x \\
\\
(0 < x \leq 1): && \int_x^1 \frac{1}{t} \d t &\leq \int_x^1 \frac{1}{t^2} \d t \\
\Rightarrow && \ln 1 - \ln x & \leq -1 + \frac{1}{x} \\
\Rightarrow && 1 - \frac1x &\leq \ln x \\
\end{align*}
\(\,\)
\begin{align*}
(1 < y): && \int_1^y \left (1 - \frac1{x} \right)\d x &\leq \int_1^y \ln x \d x \\
\Rightarrow && \left [x - \ln x \right]_1^y & \leq \left [ x \ln x - x\right]_1^y \\
\Rightarrow && y - \ln y - 1 &\leq y \ln y - y +1 \\
\Rightarrow && 2y-2 & \leq (y+1) \ln y \\
\Rightarrow && \frac{2}{y+1} & \leq \frac{\ln y}{y-1} \\
(0 < y < 1): && \int_y^1 \left (1 - \frac1{x} \right)\d x &\leq \int_y^1 \ln x \d x \\
\Rightarrow && \left [x - \ln x \right]_y^1 & \leq \left [ x \ln x - x\right]_y^1 \\
\Rightarrow && 1 - (y - \ln y) &\leq -1-(y \ln y-y) \\
\Rightarrow && 2-2y &\leq -(y+1)\ln y \\
\Rightarrow && \frac{2}{y+1} &\leq \frac{-\ln y}{1-y} \tag{\(1-y > 0\)} \\
\Rightarrow && \frac{2}{y+1} &\leq \frac{\ln y}{y-1}
\\
\\
(1 < y): && \int_1^y \ln x \d x &\leq \int_1^y (x-1) \d x \\
\Rightarrow && \left [x \ln x -x \right]_1^y &\leq \left[ \frac12 x^2 - x \right]_1^y\\
\Rightarrow && y \ln y - y +1 &\leq \frac12y^2 - y+\frac12 \\
\Rightarrow && y \ln y &\leq \frac12 \left (y^2-1 \right) \\
\Rightarrow && \frac{\ln y}{y-1} &\leq \frac{y+1}{2y} \\
\\
(0 < y < 1) && \int_y^1 \ln x \d x &\leq \int_y^1 (x-1) \d x \\
\Rightarrow && \left [x \ln x -x \right]_y^1&\leq \left[ \frac12 x^2 - x \right]_y^1\\
\Rightarrow && -1-(y \ln y - y +1) &\leq-\frac12 - \left ( \frac12y^2 - y\right)\\
\Rightarrow && \frac12 \left (y^2-1 \right) &\leq y \ln y \\
\Rightarrow && \frac{\ln y}{y-1} & \leq \frac{y+1}{2y} \tag{\(y-1 < 0\)}
\end{align*}