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Prove that for every $x>0$: $$\ln x \ge \frac{x-1}{x}$$

What I did:

$$f(x) = \ln x, \text{ } g(x) = \frac{x-1}{x} $$

$$f(1) = g(1) = 0 $$

So it's enough to prove that $$ f'(x) \ge g'(x)$$ I proved this for every $x>1$. How can I prove this for every $0<x\le1$ ?

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  • $\begingroup$ The only thing you still need to show is $f'(x) \leq g'(x)$ for $x\in (0,1]$ $\endgroup$ – Gregor de Cillia Feb 7 '16 at 17:02
  • $\begingroup$ For $0<x<1$ substitute $y=x^{-1)$ $\endgroup$ – Urgje Feb 7 '16 at 17:02
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You approach is good, but there is another which is worthy of consideration. Define $h : (0,\infty) \rightarrow \mathbb{R}$ by \begin{equation} h(x) = x \text{ln}(x) - x + 1. \end{equation} and show that this function is non-negative if and only if $f(x) \ge g(x)$. Then subject $h$ to a standard functional analysis in order to determine its range. Compute the limits as $x$ tends to $0$ and $\infty$. Moreover, the derivative of $h$ is easily seen to be $\text{ln}(x)$. You should find that $h$ is always non-negative.

Edit: word "positive" replaced with the correct word "non-negative".

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  • $\begingroup$ The limit of h as $x$ tends to 0 is $1$ and as x tends to $\infty$ is $\infty$ $h' = lnx$ indeed, But what can I conclude from all of this? $\endgroup$ – Sijaan Hallak Feb 7 '16 at 17:05
  • $\begingroup$ You will find that $h$ has only a single minimum. Specifically, at $x=1$, where the value is 0. Therefore, $h \ge 0$.I will make a single edit to correct the word positive. $\endgroup$ – Carl Christian Feb 7 '16 at 17:11
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I thought it might be instructive to present a way forward that does not rely on calculus, but rather elementary analysis only.

In THIS ANSWER and THIS ONE, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality

$$e^x\ge 1+x \tag 1$$

Setting $x=-z/(z+1)$ into $(1)$ and taking the logarithm of both sides reveals

$$\log(1+z)\ge \frac{z}{z+1} \tag 2$$

for $z>-1$. Finally, substituting $x=1+z$ in $(2)$ yields the coveted inequality

$$\frac{x-1}{x}\le \log x$$

for $x>0$.

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HINT: set $$f(x)=\ln(x)-\frac{x-1}{x}$$ then we get $$\lim_{x \to 0+}f(x)=+\infty$$ further is $$f'(x)=\frac{x-1}{x^2}$$ and $$f''(x)=-\frac{x-2}{x^3}$$ can you proceed?

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  • $\begingroup$ Cant understand what's your point here.. can you explain more? $\endgroup$ – Sijaan Hallak Feb 7 '16 at 17:04
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    $\begingroup$ show that $x=1$ gives the minimum of the given function $\endgroup$ – Dr. Sonnhard Graubner Feb 7 '16 at 17:07

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