How to give a strong proof for the given inequality based on their derivatives? This is the actual problem quoted from the book:

By considering the gradients of $ln(x)$ and $x$ show that  $ln (n)$ $<$ $n$ for $n\ge1$.

What I did:


*

*Found the gradients of both the functions in the inequality

*Compared them through inequality $\frac{1}{x} < 1$.

*As the gradient of $n$ is always greater so I thought problem is solved.


But I don't think I have a strong proof. Please give a strong proof for the problem.
 A: Hint: for $n> 1$,
$$x > 1 \implies \frac1x < 1 \implies \int_1^n \frac1xdx < \int_1^n 1\cdot dx \implies \log(n) <  n-1$$
A: If $x=1$, then $x-\log x = 1 > 0$; if $x > 1$, then $D(x-\log x) = 1 - \frac{1}{x} > 0$, implying that $x \mapsto x-\log x$ is strictly increasing on $[1, \infty[$; hence $x - \log x > 0$ for all $x \geq 1$. 
A: The technique of comparing functions is actually based on the monotonic nature of functions. For example suppose we want to prove that $f(x) > g(x)$ for all $x \geq a$. This is same as proving that $f(x) - g(x) > 0$ for all $x \geq a$. One possible approach to solve this problem is to hope that the function $h(x) = f(x) - g(x)$ is positive for $x = a$ and also increasing for all $x \geq a$. This will ensure that $h(x) > h(a) > 0$ for all $x > a$. For increasing nature of $h(x)$ we need to check the sign of derivative $h'(x)$.
Once we have understood the principle above let's us try our hand at the current question. Let $f(x) = x - \log x$ and we need to prove that $f(x) > 0$ for all $x \geq 1$. Clearly for $x = 1$ we have $f(1) = 1 - \log 1 = 1 > 0$. And $f'(x) = 1 - (1/x) = (x - 1)/x > 0$ for all $x > 1$. It follows that $f(x)$ is strictly increasing in interval $[1, \infty)$. Hence $f(x) > f(1) > 0$ for all $x > 1$. It follows that $f(x) > 0$ for all $x \geq 1$ and therefore $x > \log x $ for all $x \geq 1$.
