Inequality with Sum of Binomial Coefficients My friend noticed that for $n>12$, we have the following pattern in the binomial coefficients.

$$\sum_{i=0}^{\lfloor n/3 \rfloor}\binom{n}{i} < \binom{n}{\lfloor n/3 \rfloor + 1}$$
$$\sum_{i=0}^{\lfloor n/3 \rfloor + 1}\binom{n}{i} > \binom{n}{\lfloor n/3 \rfloor + 2}$$

We've checked these inequalities up to $n=200$ with a computer, but have not been able to come up with a proof. 
Attempt: We've attempted a asymptotic approach using Stirling approximation
$$\binom{n}{k} \approx \sqrt{\frac{n}{2\pi k (n-k)}} \frac{n^n}{k^k (n-k)^{n-k}}$$
but approximating with an integral doesn't seem to help very much, as it seems quite hard to compare
$$\int_0^{n} \sqrt{\frac{3n}{2\pi k (3n-k)}} \cdot \frac{(3n)^n}{k^k (3n-k)^{3n-k}} dx \quad \text{and} \quad  \sqrt{\frac{3}{4\pi}} \frac{3^{3n}}{2^{2n}}$$
One thing I have realized which seems important is that
$$2 \binom{3n}{n} \approx \binom{3n}{n+1} $$
However, I have not been able to work this into a proof. Any ideas?
2nd Attempt:
Using B. Mehta's linked post, in particular this inequality,
$$\sum_{i=0}^k \binom{n}{i} \leq \binom{n}{k} \frac{n-(k-1)}{n-(2k-1)}$$
subbing in $k=\lfloor\frac{n}{3}\rfloor$, we can almost get the inequality as follows
$$\sum_{i=0}^{\lfloor n/3 \rfloor} \binom{N}{i} \leq \binom{n}{\lfloor n/3 \rfloor} \frac{n-\lfloor n/3 \rfloor + 1}{n - 2 \lfloor n/3 \rfloor + 1}$$
Now we can use that $\binom{n}{k+1} = \binom{n}{k} \frac{n-k}{k+1}$ to get
$$\sum_{i=0}^{\lfloor n/3 \rfloor} \binom{N}{i} \leq \binom{n}{\lfloor n/3 \rfloor + 1}\frac{\lfloor n/3 \rfloor+1}{n-\lfloor n/3 \rfloor}\frac{n-\lfloor n/3 \rfloor + 1}{n - 2 \lfloor n/3 \rfloor + 1}$$
But unfortunately, the inequality
$$\frac{\lfloor n/3 \rfloor+1}{n-\lfloor n/3 \rfloor}\frac{n-\lfloor n/3 \rfloor + 1}{n - 2 \lfloor n/3 \rfloor + 1} \leq 1$$ does not hold for any multiples of $3$. Any other ideas?

This is partially a comment that is slightly too long. In the Math Overflow article,  we want to bound 
$$ {{N \choose k} + {N \choose k-1} + {N \choose k-2}+\dots   \over {N \choose k}} 
= {1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots} $$
The author chooses to use a geometric series as an upper bound starting with the first term. However, we can slightly delay the geometric series to get a smaller upper bound as follows:
$$ 1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots =  \\1 + \frac{k}{N-k+1} \left(1 + \frac{k-1}{N-k+2} +  \frac{(k-1)(k-2)}{N-k+2(N-k+3)} \right) + \cdots$$
Using a geometric series for an upper bound in the inner parenthesis gives us an upper bound of 
$$ \dbinom{N}{k} \left(1 + \frac{k}{N-k+1} \cdot \frac{N-k+2}{N-2k+3} \right). $$
We can easily check that for large enough $N$,
$$1 + \frac{k}{N-k+1} \cdot \frac{N-k+2}{N-2k+3} < \frac{N-(k-1)}{N-(2k-1)}. $$
Finally we can check that the steps used by the OP in the 2nd attempt portion of the question does go through if $N = 3k$ (which was an issue last time) for large enough $N$. This is just some algebraic manipulations so I won't post it here. Hopefully I haven't made an error.
