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'Climbing' printed from https://nrich.maths.org/
Thank you Andrei Lazanu from Tudor Vianu National College,
Bucharest, Romania for this solution.
To solve the problem, I plotted three functions for $0 \leq x \leq
\pi/2$. Below are the graphs of the three functions, all three
plotted on the same figure with $y_1={2x\over \pi}$ in blue, $y_2 =
\sin x$ in green and $y_3=x$ in red.
As seen from the figure, the green graph (i.e. $y = \sin x$) is
situated between the other two graphs in the range of $0 \leq x
\leq \pi/2$. For all $x$ between 0 and $\pi/2$ the point $(x, \sin
x)$ is above the point $(x, 2x/\pi)$ and hence $${2x\over \pi} \leq
\sin x.$$ In this interval the graph of $\sin x$ lies below the
line $y=x$ because this line meets the graph of $\sin x$ at the
origin and the gradient of $\sin x$ is less than 1 for all $x$.
Hence $\sin x \leq x$.
For the second part of the problem, I have to prove that: $${\tan a
\over \tan b} < {a \over b}.$$ For any $a$ and $b$ in this
interval, both $a$, $b$, $\tan a$ and $\tan b$ are positive, so
that the original inequality is equivalent to the following one:
$${\tan a \over a} < {\tan b \over b}$$ In the figure below the
graph of function $y=\tan x/x$ is plotted for $0 < x <
\pi/2$.
I observe that the function is monotonically increasing. This means
that for any $a < b$ (but remaining in the interval $0 < a
< b < \pi/2$), $y(a) < y(b)$ as required.
Alternatively, from the graph of $y = \tan x$ in this interval, if
$a < b$ then the gradient of line joining the origin the point
$(a, \tan a)$ is less than the gradient of the line joining the
origin to the point $(b, \tan b)$ so $${\tan a \over a} < {\tan
b \over b}$$ and hence $${\tan a \over \tan b} < {a \over
b}.$$