Use the definition of $\ln(x)$ as an integral to show that $f(x)=\frac{ln(x)}{x^2} \leq 1/x$ for all $x\geq 1$. As the title says, if we let $$f(x)=\frac{ln(x)}{x^2}$$
I know that $$ln(x)=\int_1^x \frac{dt}{t} dt$$
Since $x^2 >0$ we can rewrite the question as $ln(x) \lt x$ $\forall$ $x\ge1$
How do we approach this? The one hint we have is to bound the integral with an upper sum, but that's just confusing me further.
Is it as simple as stating that the definition itself implies that $$\frac{1}{x}=\int_1^x \frac{dt}{t} dt$$ and that $$\frac{1}{x} \lt x$$
 A: One has $$\log x = \int_1^x \frac{dt}{t} \leq \int_1^x dt = x-1 < x$$ Since $x\geq 1$, dividing by $x^2$ preserves the inequality.  
A: Hint: There's a fundamental inequality involving integrals that says
$$\left | \int_a^b f(t)\, dt \right | \leq (b-a) \cdot \max\{\left | f(t) \right |\,; \, t \in (a,b)\}$$
Can you think of how to use it?
A: Note that if $1\le t\le x$, then $\frac{1}{x}\le\frac{1}{t}\le 1$.  
Therefore, we find that 
$$\frac{x-1}{x}\le \int_1^x \frac{1}{t}\,dt \le x-1 \tag 1$$
In fact, $(1)$ is true for all $x>0$.  
Finally, we have
$$\frac{x-1}{x^3}\le \frac{1}{x^2}\int_1^x \frac{1}{t}\,dt\le \frac{x-1}{x^2}<\frac1x$$
for all $x>0$
A: I find it easier to write
$\ln(1+x)
=\int_1^{1+x} \dfrac{dt}{t}
$
so the inequalities become
$\dfrac{x}{1+x}
\le \ln(1+x)
\le x
$
or,
putting 
$x-1$ for $x$,
$\dfrac{x-1}{x}
\le\ln(x)
\le x-1
$.
Dividing by
$x^2
$,
$\dfrac{x-1}{x^3}
\le \dfrac{\ln(x)}{x^2}
\le \dfrac{x-1}{x^2}
\lt \dfrac1{x}
$.
Note that
this would probably be used
for $x$ near $1$,
so there is no worry
about the lower bound
blowing up.
