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0 votes
1 answer
34 views

How to Derive the Binomial Coefficient Upper Bound and Final Inequality in "Scheduling Multithreaded Computations by Work Stealing"?

In the paper Scheduling Multithreaded Computations by Work Stealing under the section "Atomic accesses and the recycling game", it mentions the binomial coefficient approximation: $$ \binom{...
grzhan's user avatar
  • 3
5 votes
3 answers
132 views

Strengthening the Log-Concavity of Binomial Coefficients: $\binom{n}{k-1}\binom{n}{k+1} < \binom{n}{k}^2 - \binom{n}{k}$

In the following question: Log concavity of binomial coefficients: $ \binom{n}{k}^2 \geq \binom{n}{k-1}\binom{n}{k+1} $ It is proven via a combinatorial injective argument. However, by noticing that ...
Jack Tiger Lam's user avatar
2 votes
0 answers
132 views

how to prove $\sum_{j=\frac{k}{2}+1}^{k-1}\binom{k-1}{j}\binom{n-(k-1)}{k-j}<2\binom{n-1}{k-1}$?

I want to prove for $2\le k<n$ (and $k$ is even if necessary) $$\sum_{j=\frac{k}{2}+1}^{k-1}\binom{k-1}{j}\binom{n-(k-1)}{k-j}<2\binom{n-1}{k-1}.$$ MATLAB verifies it is true when $n=30$ and $k=...
Connor's user avatar
  • 2,075
0 votes
1 answer
56 views

How to get this lower bound for $C(m^2,m-1)$?

While applying Nechiporuk’s Theorem to get a lower bound on the formula size of Element distinction function, the following inequality is used: (ref here) $C(m^2,m-1) \geq (m^2-m+1)^{m-1}$. Is there a ...
emmy's user avatar
  • 214
3 votes
1 answer
110 views

Alternative proof for ${pn \choose n}{qn \choose n} \ge {pqn \choose n}$ inequality

Reading this question, I saw that for $p,q,n$ positive integers the following inequality holds: $${pn \choose n}{qn \choose n} \ge {pqn \choose n}$$ The inequality is not tight. A simple combinatorial ...
Fabius Wiesner's user avatar
3 votes
3 answers
313 views

Good upper bound on ratio of binomial coefficients

What is the asymptotic behavior of $n/2+k/2 \choose k$/$n\choose k$ for $k\le n/2$. Can we show that it goes down (sub)exponential in $k$ i.e. $n/2+k/2 \choose k$/$n\choose k$ $\le c^{-k^{\epsilon}}$ ...
user1131274's user avatar
  • 1,693
2 votes
0 answers
81 views

Inequality of Narayana numbers

In some computations I am making, it is desirable to deduce that a certain expression is positive. I managed to rewrite it in terms of Narayana numbers, which are defined by $N(n,i) := \frac{1}{n}\...
Luis Ferroni's user avatar
3 votes
3 answers
126 views

Prove that $\binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m$ for $m\in \mathbb{N}\backslash \{0,1\}$

Prove the following inequality for every $m\in \mathbb{N}\backslash \{0,1\}$: $$ \binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m.$$ By some computational arguments, the inequality seems to be true and in ...
Hola's user avatar
  • 185
1 vote
1 answer
56 views

proving an inequality based on double products and binomial

Iv been trying to prove the following inequality: let $a_1,\ldots,a_n$ be a non-increasing sequence, i.e., $a_1\geq a_2\geq \cdots \geq a_n\geq 0$ such that $\sum_i a_i=m$. Then, prove that $$\prod_{i=...
Vaas's user avatar
  • 413
0 votes
0 answers
53 views

Hints and tips to prove this inequality.

Let's say that $n = \lfloor{2^{\frac{k}{2}}}\rfloor$ for some integer $k \geq 3$. I need to prove this inequality: $${n \choose k}\cdot2^{1-{k \choose 2}} < 1$$ I tried to rewrite those terms or ...
User1234's user avatar
4 votes
1 answer
69 views

For $n \geq 2$, show that $\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} < \sqrt{2 ^ {n - 1} n ^ 3}$

Good day, Can someone help me with giving hints for this problem: Show that for $n \geq 2, n \in \mathbb{Z}$, $$\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} < \sqrt{2 ^ {n - 1} n ^ 3}$$ I tried $$\sum_{...
MangoPizza's user avatar
  • 1,858
1 vote
0 answers
92 views

Summing binomial coefficients and manipulating summation order

I'm having trouble understanding Example 3 (pg. 61). As a preliminary, the solution makes use of the following two identities, which I've restated below for convenience: Eq. (25) on pg. 59: $$\sum_{k=...
user51462's user avatar
  • 673
5 votes
2 answers
258 views

More elegant proofs of $\binom a2+\binom b2\leq \binom{a+b-1}2$

I came to know about the inequality $$\binom a2+\binom b2\leq \binom{a+b-1}2$$ and tried to prove it. It was quite easy to derive it using brute force algebraic calculations. All boils down to showing ...
Sayan Dutta's user avatar
  • 9,592
3 votes
1 answer
129 views

Where does this inequality come from?

In Bollobás's Modern Graph Theory, section IV.4, Page 120 line -10, he uses the inequality $$\frac{t {n \choose t}}{n {\epsilon n \choose t}} \leq \frac{t}{n} \epsilon^{-t} (1-\frac{t}{\epsilon n})^{-...
CTVK's user avatar
  • 467
13 votes
1 answer
614 views

Inequality between binomial sums

I want to prove that the following inequality holds whenever $k\leq n-1$ and $1\leq i\leq \lfloor\frac{k}{2}\rfloor$. $$\frac{2}{C}\binom{n}{k}\binom{k}{i+1} \leq \frac{k}{i}\sum_{j=i}^{k-1} (k-j)\...
Luis Ferroni's user avatar

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