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I have come across an inequality involving factorials and exponentials in many different forms. How do I go about proving this inequality?

$$\sum_{t=k}^n \binom {n}{t}\binom {\binom {t}{k}}{\frac{2t}{k}}q^{-2t/k} <= \sum_{t=k}^n \Biggl[\frac{en}{t}\Biggl(\frac {ek\binom tk}{2tq}\Biggl) ^ {2/k}\Biggl] ^ t $$

Thank you.

ondrejba
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  • Have you tried comparing the individual terms? Also, where did you see this? – marty cohen Oct 10 '17 at 21:13
  • I had tried to solve the problem for several hours before M. Lavrov helped me. I found this inequality in Waiter-Client and Client-Waiter colourability games on a k-uniform hypergraph and the k-SAT game - W.E. Tan (2016). – ondrejba Nov 18 '17 at 18:00

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The key is the inequality $$\left(\frac nk\right)^k \le \binom nk \le \left(\frac{en}{k}\right)^k$$ which we use for turning binomial coefficients into expressions that are easy to do algebra with. (See, e.g., this question for proof.) Actually, here we only use the upper bound, but it's good to keep both of them in mind so that you know how far it's off from the truth.

Here, once we write $\binom nt \le \left(\frac {en}{t}\right)^t$ and $\binom{\binom nk}{2t/k} \le \left(\frac{e\binom nk}{2t/k}\right)^{2t/k}$, we get the expression $$\sum_{t=k}^n \binom {n}{t}\binom {\binom {t}{k}}{\frac{2t}{k}}q^{-2t/k} \le \sum_{t=k}^n \left(\frac{en}{t}\right)^t \left(\frac{e k \binom nk}{2t}\right)^{2t/k} q^{-2t/k}$$ and we realize that all factors are raised to the $t^{\text{th}}$ power, so we can rewrite them as $$\sum_{t=k}^n \left[ \frac{en}{t} \left(\frac{ek\binom nk}{2t}\right)^{2/k} q^{-2/k}\right]^t.$$ Similarly, we realize that we can combine the last two factors, because they are both raised to the $2/k^{\text{th}}$ power.

Misha Lavrov
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