# A combinatorial inequality $\binom {2n}{k} + \binom {2n}{n-k} \ < \ \binom{2n}{n} + \binom{2n}{0} \ \ \text{for} \ \ k<n \ \$

I am struggling some math problems.

Fighting some problems, I find out a rule.



Could you please see the table below?

HERE is my question!

I (may) found out the inequality



$$\binom {2n}{k} + \binom {2n}{n-k} \ < \ \binom{2n}{n} + \binom{2n}{0} \ \ \text{for} \ \ k

or

$$\binom {2n}{k} + \binom {2n}{n-k} \ < \ \binom{2n}{l} + \binom{2n}{n-l} \ \ \text{for} \ \ k

$$\text{Actually (2) implies (1)}$$ 

1. Is these inequality exist already?
2. Is these inequality true, INTUITIVELY?
3. Is these inequality TRUE?
4. If these are right, then how can I proof?

 (I wanna believe these are true, and then these will proof simply...)

Thank you for your attention to this matter.

• It fails for $k=0$. – lhf Oct 6 '18 at 16:19
• The inequality $(1)$ does not hold for $(n,k)=(2,1),(3,1),(3,2)$. – mathlove Oct 7 '18 at 3:52