Prove inequalities of the log of 1 + 1/x

For $x > 0$ prove that

$\frac{1}{x + \frac{1}{2}} < \log \left( 1 + \frac{1}{x} \right) < \frac{1}{x}.$

Before the proof, a picture for this one is probably useful. In the graph below we are going to get the inequality on the right by comparing the area under the graph of $\frac{1}{x}$ from $x$ to $x+1$ (blue curve) and the area of the rectangle under the blue dashed line from $x$ to $x+1$ . We’ll show that the area of the rectangle is $1/x$ and the area under the curve is $\log \left( 1 + \frac{1}{x} \right)$ .

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Proof. First, we note that

$\log \left( 1 + \frac{1}{x} \right) = \log \left( \frac{x+1}{x} \right) = \log (x+1) - \log x.$

Furthermore,

$\int_x^{x+1} \frac{1}{t} \, dt = \log (x+1) - \log x.$

Therefore we have

$\log \left( 1 + \frac{1}{x} \right) = \int_x^{x+1} \frac{1}{t} \, dt.$

(In the picture this is the area under the curve between $x$ and $x+1$ .)
Since the function $\frac{1}{t}$ is strictly decreasing on the positive real axis (since it’s derivative is $-\frac{1}{t^2}$ is negative everywhere) we know that for any fixed $x> 0$ we have $\frac{1}{x} > \frac{1}{t}$ for every $t \in (x, x+1)$ . Therefore, by the monotone property of the integral,

$\int_x^{x+1} \frac{1}{t} \, dt < \int_x^{x+1} \frac{1}{x} \, dt = \frac{1}{x} \int_x^{x+1} dt = \frac{1}{x}.$

(Note the $\frac{1}{x}$ inside the integral is a constant here since we have chosen some fixed $x>0$ . For any fixed $x$ the point $\frac{1}{x}$ is the point on the curve $f(x)$ at the left end of the interval. So, this is saying the integral of the curve $\frac{1}{t}$ from $x$ to $x+1$ is less than the integral of the rectangle of height $\frac{1}{x}$ from $x$ to $x+1$ .) This gives us the inequality on the right that we wanted,

$\log \left( 1 + \frac{1}{x} \right) < \frac{1}{x}.$

Now, to prove the inequality on the left we know that $x > 0$ if and only if $\frac{1}{x} > 0$ . Thus, the given inequality holds for all $x>0$ if and only if it holds for all $\frac{1}{x} > 0$ . Therefore,

$\frac{1}{x+\frac{1}{2}} < \log \left( 1+ \frac{1}{x} \right)$

if and only if

$\frac{1}{\frac{1}{x} + \frac{1}{2}} < \log \left( 1 + \frac{1}{1/x} \right) \implies \frac{x}{1+\frac{1}{2} x} < \log (1+x).$

Now, consider

$\begin{align*} f(x) &= \log(1+x) - \frac{x}{1+\frac{1}{2}x} \\[9pt] f'(x) &= \frac{1}{1+x} - \frac{1+\frac{1}{2}x - \frac{1}{2}x}{(1+\frac{1}{2}x)^2} \\[9pt] &= \frac{1}{1+x} - \frac{1}{1+x+\frac{1}{4}x^2} \\[9pt] &> 0 \end{align*}$

for all $x > 0$ . Therefore, $f(x)$ is increasing on the positive real axis. Since

$f(0) = \log (1+0) - \frac{0}{1+\frac{1}{2}\cdot0} = 0,$

we have that $f(x) > 0$ for all $x > 0$ . Hence, we obtain the inequality on the left

$\frac{1}{x+\frac{1}{2}} < \log \left( 1+\frac{1}{x} \right). \qquad \blacksquare$

Stumbling Robot

Prove inequalities of the log of 1 + 1/x

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