I want to prove that the binomial coefficient ${n \choose k}$ for $n \ge k$ is a monotonically nondecreasing sequence for a fixed $k$. How do I do this?

This question already has an answer here: Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ 7 answers

Given: $0<x<y<1$ $z=x+y$ $x=u$ $y=z-u$ $0<u<z-u<1$ I need to show that this implies: If $0<z<1$, then $0<u<\frac{z}{2}$, and If $1<z<2$, then $z-1<u<\frac{z}{2}$. I can observe that $0<z<2$ but I am not able to make any progress on getting the 2 required implications. Please show me the steps as well as the reasoning to solve these […]

Only a hint on how to prove this, if not a complete proof, would also be appreciated.

Let $(X,\mathbb X,\mu)$ be a measure space and let $(E_n)$ be a sequence in $\mathbb X$. Show that $$\mu(\lim\inf E_n)\leq\lim\inf\mu(E_n).$$ I am quite sure I need to use the following lemma. Lemma. Let $\mu$ be a measure defined on a $\sigma$-algebra $\mathbb X$. If $(E_n)$ is an increasing sequence in $\mathbb X$, then $$\mu\left(\bigcup_{n=1}^\infty E_n\right)=\lim\mu(E_n).$$ […]

Today I’ve seen a question like this: $$\text{Given } x^2+y^2 \leq 16, \text{ what is the maximum value for } 3x+4y?$$. What I’ve tried was the following: $$3x+4y \leq \sqrt{x^2+y^2}\cdot\sqrt{9+16}$$ But the problem here is that I have to give a value for $\sqrt{x^2 +y^2}$ and I know that this would not give me a […]

let $a_{1},a_{2},\cdots,a_{n}\ge 0$,and such $a_{1}+a_{2}+\cdots+a_{n}=1$. Find this follow minimum $$I=a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}$$ My try:since $$a^2_{1}+a^2_{2}+\cdots+a^2_{n}\ge a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1}$$ this is true because $$\Longleftrightarrow \dfrac{1}{2}[(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+\cdots+(a_{n-1}-a_{n})^2+(a_{n}-a_{1})^2]\ge0$$ so \begin{align*} &a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}\\ &\ge a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2(a^2_{1}+a^2_{2}+\cdots+a^2_{n})\\ &=-(a^2_{1}+a^2_{2}+\cdots+a^2_{n}) \end{align*} if we use Cauchy-Schwarz inequality,we have $$(a^2_{1}+a^2_{2}+\cdots+a^2_{n})(1+1+\cdots+1)\ge (a_{1}+a_{2}+\cdots+a_{n})^2=1$$ But this is not usefull. and Then I can’t, yesterday I have ask this problem：How find this inequality minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$, […]

I was given a task to prove that inequality is true for x>0: $(e^x-1)\ln(1+x) > x^2$. I’ve tried to use derivatives to show that the $f(x) = (e^x-1)\ln(1+x)-x^2$ is greater than zero, but has never succeeded. Any help will be appreciated.

Nice Problem: Let $n\ge 3$,and $y_{1},y_{2},\cdots,y_{n}$ be real numbers, and such that $$2y_{k+1}\le y_{k}+y_{k+2}.1<k\le n-2$$ and $\displaystyle\sum_{k=1}^{n}y_{k}=0$. Show that $$\sum_{k=1}^{n}k^2y_{k}\ge (n+1)\sum_{k=1}^{n}ky_{k}.$$ I have understand this following nice Proof from AMM11866: Note \begin{align*} &0\le\sum_{k=1}^{n-2}\binom{k+1}{2}\binom{n-k}{2}(y_{k+2}-2y_{k+1}+y_{k})\\ &=\sum_{k=1}^{n}\left(\binom{k+1}{2}\binom{n-k}{2}-2\binom{k}{2}\binom{n-k+1}{2}+\binom{k-1}{2}\binom{n-k+2}{2}\right)y_{k}\\ &=\sum_{k=1}^{n}\left(3k^2-3(n+1)k+\binom{n+2}{2}\right)y_{k}\\ &=3\left(\sum_{k=1}^{n}k^2y_{k}-(n+1)\sum_{k=1}^{n}ky_{k}\right) \end{align*} Question 1: I want to know this proof with the coefficient $\binom{k+1}{2}\binom{n-k}{2}$. How to get it? Question 2: […]

Let $P_{0},P_{1},P_{2},\cdots,P_{n}$ be $n+1$ points in the plane. Let $ d=1$ denote the minimal value of all the distances between any two points. Prove that $$\dfrac{1}{P_{0}P_{1}}+\dfrac{1}{P_{0}P_{2}}+\cdots+\dfrac{1}{P_{0}P_{n}}<\sqrt{15n}$$ This problem background is from China high school math competition (Oct 14, 2012) problem 15,can see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2822547&sid=bbbc81f99da00d657f61b4835931c87e#p2822547 also can see this two solution:http://wenku.baidu.com/view/82fb84d4240c844769eaeea3.html But for my problem,I can’t prove […]

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