The sum of binomial coefficients up to $k\le n/3$ does not exceed the $k$th coefficient

Question

How would you prove the following (for when $k\leq\frac{n}{3}$)?

$$\sum_{i=0}^{k-1} \binom ni \le \binom nk$$

Answer 1

Consider that: $$\frac{\binom{n}{k-1}}{\binom{n}{k}}=\frac{k}{n-k},\quad\frac{\binom{n}{k-2}}{\binom{n}{k}}=\frac{k-1}{n+1-k}\cdot\frac{k}{n-k}\leq\left(\frac{k}{n-k}\right)^2,\quad \frac{\binom{n}{k-i}}{\binom{n}{k}}\leq\left(\frac{k}{n-k}\right)^i$$ so: $$\frac{1}{\binom{n}{k}}\sum_{i=0}^{k-1}\binom{n}{i}\leq\sum_{j=1}^{k}\left(\frac{k}{n-k}\right)^j\leq\frac{1}{3}+\frac{1}{9}+\ldots+\frac{1}{3^k}<\frac{1}{2}.$$ A similar proof works even if $k\leq n/3$: $$\frac{1}{\binom{n}{k}}\sum_{i=0}^{k-1}\binom{n}{i}\leq\sum_{j=1}^{k}\left(\frac{k}{n-k}\right)^j\leq\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^k}<1.$$