Bounds for the exponential integral In  Abramowitz and Stegun: Handbook of Mathematical Functions
(on page 229, property 5.1.20) it is found that
$$
\frac{1}{2} \log \left(1 + \frac{2}{x} \right) < \exp(x) E_1(x) < \log \left(1 + \frac{1}{x} \right) \qquad (x > 0)
$$
where 
$$
E_1(x) = \int_x^\infty \frac{\exp(-t)}{t} dt
$$
How does the proof go?
 A: Here is a proof following from the theorem that, if for all $x$, $g'(x) < f'(x) < h'(x)$, and the inequality $g(x)<f(x)<h(x)$ holds at some point, then this inequality holds true for all $x$.
First we can rearrange the inequality as
$$e^{-x}\frac{\ln(x+2)-\ln(x)}{2}< \int_x^\infty \frac{e^{-t}}{t}dt<e^{-x}  \big(\ln(x+1)-\ln(x)\big)$$
Then we can take the derivative of each side and divide by $e^{-t}$ to obtain another inequality to prove.
$$\frac{\ln(x)-\ln(x+2) +\frac{1}{x+2} - \frac{1}{x}}{2} <\frac{-1}{x}, < \ln(x)-\ln(x+1) +\frac{1}{x+1}-\frac{1}{x}$$
Now we can continue down the line, taking the derivative again.
$$\frac{\frac{1}{x}-\frac{1}{x+2}-\frac{1}{(x+2)^2}+\frac{1}{x^2}}{2} < \frac{1}{x^2} < \frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2}+\frac{1}{x^2}$$Finally, we have reduced the problem to something easier to prove. For the LHS, we can multiply out $2x^2(x+2)^2$ (a positive value since $x>0$) to obtain the equivalent inequality,
$$x(x+2)^2-x^2(x+2) - x^2 + (x+2)^2<2(x+2)^2$$
Now we can expand and everything cancels out nicely to get
$$x^3+4x^2+4x-x^3-2x^2 - x^2-x^2-4x-4 = -4<0$$
Now for the RHS, we can similarly multiply by $x^2(x+1)^2$ to obtain another equivalent inequality
$$(x+1)^2<x(x+1)^2-x^2(x+1) -x^2+(x+1)^2$$
Expanding and canceling we get
$$0<x^3+2x^2+x-x^3-x^2-x^2=x$$
Which is true since we are only considering $x>0$. Now that we've finally proved that $g''(x)<f''(x)<h''(x)$, the last step is to show that $g'(x)<f'(x)<h'(x)$ and $g(x)<f(x)<h(x)$. These can be verified at $x=1$, and if anyone would like more on this verification, just ask.
