International Mathematical Olympiad 51 Selection Test

Lemma. For all integers $0 \leq t, s \leq m$ such that $t + s \leq m$, $\frac{\binom{2m}{m-s}}{\binom{2m}{m-t-s}} > \frac{t^2}{m}$.
Proof of lemma.
$$A = \frac{\binom{2m}{m-s}}{\binom{2m}{m-t-s}} = \frac{(m-t-s)!(m+t+s)!}{(m-s)!(m+s)!} =$$
$$= \left( \frac{(m+t+1)(m+t+2)\dots(m+t+s)}{(m-t-s+1)(m-t-s+2)\dots(m-s)} \right) = \\ = \left( 1 + \frac{t+2s}{m-t-s+1} \right) \left( 1 - \frac{t+2s}{m-t-s+2} \right) \dots \left( 1 + \frac{t+2s}{m-s} \right).$$
Using the $(m-t-s+1) < (m-t-s+2) < \dots < (m-s)$ and Bernoulli's inequality:
$$A \geq \left(1 + \frac{t+2s}{m-s}\right)^t > 1 + \left(\frac{t+2s}{m-s}\right)t > \frac{t^2}{m}.$$
This completes the proof of lemma.

Let us compare the sum
$$\binom{2m}{0} + \binom{2m}{1} + \dots + \binom{2m}{m-t-1}$$
with the sum
$$\binom{2m}{t} + \binom{2m}{t+1} + \dots + \binom{2m}{m-1} \quad (1).$$
Sum in (1) is clearly less than $\frac{2^{2m}}{2}$. $t \leq [\sqrt{km}]$ implies $\frac{t^2}{m} \leq k$ and we have $\binom{2m}{m-t-1} \leq \frac{1}{2}\binom{2m}{m-1}$ by lemma. Similarly, $\binom{2m}{m-t-2} \leq \frac{1}{k}\binom{2m}{m-2}$, $\dots$, $\binom{2m}{0}\frac{1}{k}\binom{2m}{t}$.
Hence we get that,
$$\begin{aligned} & \binom{2m}{0} + \binom{2m}{1} + \dots + \binom{2m}{m-t-1} < \\ & \frac{1}{k} \left( \binom{2m}{t} + \binom{2m}{0} + \binom{2m}{1} + \dots + \binom{2m}{m-t-1} \right) < \\ & \frac{1}{k} \left( \binom{2m}{t} + \binom{2m}{t+1} + \dots + \binom{2m}{m-1} \right) < \frac{2^{2m}}{2k}. \end{aligned}$$