Step i 1990 question 9 solution

\displaystyle y = \frac, \; \frac{\mathrmy}{\mathrmx} = -\frac}

Therefore the equation of the tangent at the point \displaystyle \left(b, \frac\right) is \displaystyle y = -\frac}x + \frac

The co-ordinates of C are at the intersection of \displaystyle y = -x + 2 and \displaystyle y = -\frac} + \frac

\displaystyle -x + 2 = -\frac} + \frac \Rightarrow x\left(1-\frac}\right) = 2\left(1-\frac\right)

\displaystyle x = \frac} = \frac

\displaystyle y = \frac

Using the expression for the area of a trapezium:

Area of ACC'A' = \displaystyle \frac\left(1 + \frac\right)\left(\frac\right) = \frac{(b+1)^}

Area of CBB'C' = \displaystyle \frac\left(\frac + \frac\right)\left(b - \frac\right) = \frac{(3b+1)(b-1)}}

Sum of Areas = \displaystyle \frac} = \frac + 4b - 1}} = \frac

As the gradient is constantly increasing (becoming 'less' negative), the area under the curve between 1 and b is greater than the sum of the areas of ACC'A' and CBB'C:

\displaystyle \int^_ \frac \; \mathrmx = \ln b \geq \frac, \; b > 1

\displaystyle \Rightarrow \ln (1 + z) \geq \frac, \; z > 0

To prove the left-hand inequality, it is sufficient to show that \displaystyle \frac \geq \frac, \; z > 0

\displaystyle 5z^ + 14z + 8 \geq 4z^ \Leftrightarrow z^ + 14z + 8 \geq 0, \; z > 0

Which is clearly true.

The inequality \displaystyle \ln(1+z) \leq z follows immediately from the fact that 1 + z \leq e^ by the Maclaurin expansion of e^.

Alternatively, note that the area of the trapezium ABB'A' is greater than the area under the curve between 1 and b.

\displaystyle \ln b \leq \frac\left(1 + \frac\right)\left(b-1 \right), \; b > 1

\displaystyle \ln (1+z) \leq \frac + 2z}, \; z>0

\displaystyle \frac + 2z} \leq z \Leftrightarrow z^ + 2z \leq 2z^ + 2z \Leftrightarrow 0 \leq z^

Therefore \displaystyle \frac \leq \ln (1+z) \leq z, \; z>0

Solution by Dystopia.