For prove that

Before the proof, a picture for this one is probably useful. In the graph below we are going to get the inequality on the right by comparing the area under the graph of from to (blue curve) and the area of the rectangle under the blue dashed line from to . We’ll show that the area of the rectangle is and the area under the curve is .

*Proof.* First, we note that

Furthermore,

Therefore we have

(In the picture this is the area under the curve between and .)

Since the function is strictly decreasing on the positive real axis (since it’s derivative is is negative everywhere) we know that for any fixed we have for every . Therefore, by the monotone property of the integral,

(Note the inside the integral is a constant here since we have chosen some fixed . For any fixed the point is the point on the curve at the left end of the interval. So, this is saying the integral of the curve from to is less than the integral of the rectangle of height from to .) This gives us the inequality on the right that we wanted,

Now, to prove the inequality on the left we know that if and only if . Thus, the given inequality holds for all if and only if it holds for all . Therefore,

if and only if

Now, consider

for all . Therefore, is increasing on the positive real axis. Since

we have that for all . Hence, we obtain the inequality on the left