China Hong Kong Mathematical Olympiad

a.
We label the inequality as follows.
$$g(ty + (1-t)z) \leq t g(y) + (1-t) g(z) \quad (1)$$
Putting $y = n - 1$ and $z = n + c$ in (1), we obtain
$$g(t(n-1) + (1-t)(n+c)) \leq t g(n-1) + (1-t) g(n+c).$$
Suppose $c \in [0, 1]$. Replacing $t$ by $\frac{c}{1+c} \in [0, 1]$, this becomes
$$g\left(\frac{c(n-1) + (n+c)}{1+c}\right) \leq \frac{c}{1+c} g(n-1) + \frac{1}{1+c} g(n+c).$$
Note that the left-hand side is just $g(n)$. Therefore, this is the same as
$$c [g(n) - g(n-1)] \leq g(n+c) - g(n).$$
This is exactly the inequality on the left to be proved.
For the other inequality, we simply put $y = n+1$ and $z = n$ in (1). This gives
$$g(t(n+1) + (1-t)n) \leq t g(n+1) + (1-t) g(n).$$
Note that the left-hand side is just $g(n+t)$. This is exactly the other desired inequality.

b.
Since $g(x) = \log f(x)$, the inequality in part (a) can be rewritten as
$$t (\log f(n) - \log f(n-1)) \leq \log f(n+t) - \log f(n) \leq t (\log f(n+1) - \log f(n)).$$
This implies
$$\left( \frac{f(n)}{f(n-1)} \right)^t \leq \frac{f(n+t)}{f(n)} \leq \left( \frac{f(n+1)}{f(n)} \right)^t.$$
As $f(n) = (n-1) f(n-1)$ and $f(n+1) = n f(n)$, this can be reduced to
$$(n-1)^t \leq \frac{f(n+t)}{f(n)} \leq n^t. \quad (2)$$

Now, note that $f(2) = 1 \cdot f(1) = 1$. Putting $n = 2$ and $t = \frac{1}{2}$ in (2), we obtain
$$1 \leq f\left(\frac{5}{2}\right) \leq \sqrt{2}.$$
Using $f\left(\frac{5}{2}\right) = \frac{3}{2} f\left(\frac{3}{2}\right) = \frac{3}{4} f\left(\frac{1}{2}\right)$, we get
$$$$
$\frac{4}{3} \leq f\left(\frac{1}{2}\right) \leq \frac{4}{3} \sqrt{2}.$