During my Fall 2010 ECE 516 (Information Theory) class, the instructor asked the class to prove this seemingly trivial logarithmic inequality [1] by as many different methods and, if possible, using an information theoretic result. The given inequality comes handy to prove the Gibbs Inequality [2] [3].I could think of the following ways though the desired proof still eludes me. I would present my approach to it though:

Using series expansion:
For $\forall y$, the power series expansion of the function $e^{y}$ is given by

$e^{y} = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots$

$\Rightarrow e^{y} \geq 1 + y, y \geq 0$

$\Rightarrow z \geq 1 + \ln{z}, e^y = z > 0$

$\Rightarrow -\ln{z} \geq 1 - z, z > 0$

$\Rightarrow \ln{\frac{1}{z}} \geq 1 - z, z > 0$

$\Rightarrow \ln{x} \geq 1 - \frac{1}{x}, x > 0$ and where $x = \frac{1}{z}$
Q.E.D.

Using definite integral:
For $x \geq 1$, we have area under the monotonically increasing logarithmic curve lying above the x-axis given by
$\displaystyle\int_{1}^{x}\ln{x}dx \geq 0, x\geq 1$

$\Rightarrow (x\ln{x} - x)|_{1}^{x} \geq 0, x\geq 1$

$\Rightarrow (x\ln{x} - x) - (0-1) \geq 0, x\geq 1$

$\Rightarrow (x\ln{x} - x) + 1 \geq 0, x\geq 1$

$\Rightarrow (x\ln{x} - x) \geq -1, x\geq 1$

$\Rightarrow \ln{x} \geq \frac{x-1}{x}, x\geq 1$

$\Rightarrow \ln{x} \geq 1-\frac{1}{x}, x\geq 1$ … (1)

For $0 < x \leq 1$, we have area under the monotonically increasing logarithmic curve lying below the x-axis given by

$\displaystyle\int_{x}^{1}\ln{x}dx \leq 0, 0 < x \leq 1$

$\Rightarrow \displaystyle\int_{1}^{x}\ln{x}dx \geq 0, 0 < x \leq 1$

Following similar steps as above for $x \geq 1$, we get,
$\Rightarrow \ln{x} \geq 1-\frac{1}{x}, 0 < x \leq 1$ … (2)

Result follows from (1) and (2).
Q.E.D.

Using the definition of convexity:
Let’s assume $f(x) = \ln{x}-1+\frac{1}{x}$ where $x>0,$ and then check the convexity of this function:

$\frac{df(x)}{dx} = \frac{1}{x} - \frac{1}{x^2} \mbox{ equated to zero gives,}$

$x(x-1) = 0$

$\Rightarrow x = 1 \mbox{ is the solution. } (\because x>0)$

$\mbox{Now, } \frac{d^2f(x)}{dx^2}| = (\frac{-1}{x^2} + \frac{2}{x^3})$

$\mbox{For x = 1, } \frac{d^2f(x)}{dx^2}|_{x=1} = -1 + 2 = 1 \mbox{, which is positive}$

$\Rightarrow \mbox{by definition of convexity, f(x) is convex at x = 1 for x\textgreater0}$

$\Rightarrow \mbox{for any x \textgreater 0, } f(x) \geq f(1)$

$\Rightarrow \mbox{for any x \textgreater 0, } \ln{x}-1+\frac{1}{x} \geq 0$

$\Rightarrow \mbox{for any x\textgreater 0, } \ln{x} \geq 1-\frac{1}{x}$
Q.E.D.

Using an information theoretic result:
This is just a vaporware [4] as of now. However, as shown in [2], this logarithmic inequality can be used to prove Jensen’s Inequality [5] or properties of Kullback-Leibler distance [6]. So is it possible that one of these results can be employed to do the inverse i.e. to prove the above-mentioned logarithmic inequality?

References:
[1] Cover T. M. and Thomas J. A., “Elements of Information Theory,” Wiley-Interscience, 2nd Edition, 2006, Problem 2.13.
[2] Gibbs’ Inequality.
[3] Proof of Gibbs’ Inequality. Please note that this document misspells the inequality as Gibb’s.
[4] Vaporware.
[5] Jensen’s Inequality.
[6] Kullback-Leibler Distance.