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?