Articles of inequality

How prove this mathematical analysis by zorich from page 233

Let $f$ be twice differentiable on an interval $I$,Let $$M_{0}=\sup_{x\in I}{|f(x)|},M_{1}=\sup_{x\in I}{|f'(x)|},M_{2}=\sup_{x\in I}{|f”(x)|}$$ show that (a):$$M_{1}\le 2\sqrt{M_{0}M_{2}}$$ if the length of $I$ is not less than $2\sqrt{\dfrac{M_{0}}{M_{2}}}$ (b):the numbers $2$ and $\sqrt{2}$ (in part a) cannot be replaced by smaller numbers. My try:for part $(a)$ I can prove if the length of $I$ is not […]

System of Equations: any solutions at all?

I am looking for any complex number solutions to the system of equations: $$\begin{align} |a|^2+|b|^2+|c|^2&=\frac13 \\ \bar{a}b+a\bar{c}+\bar{b}c&=\frac16 (2+\sqrt{3}i). \end{align}$$ Note I put inequality in the tags as I imagine it is an inequality that shows that this has no solutions (as I suspect is the case). This is connected to my other question… I have […]

Proving $\frac{x}{x^2+1}\leq \arctan(x)$ for $x\in .$

How can I prove this inequality? $$\frac{x}{x^2+1}\leq \arctan(x) \, , x\in [0,1].$$ Thank you so much for tips! Sorry if this is just stupid.

How can I prove that the binomial coefficient ${n \choose k}$ is monotonically nondecreasing for $n \ge k$?

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?

Inequality and Induction: $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$

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

Help demonstrate how to arrive at the implication of some given inequalities and equations

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 […]

How do I prove $\sqrt{x^2 + y^2} \le |x| + |y|$?

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

An inequality about sequences in a $\sigma$-algebra

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 $$\mu\left(\bigcup_{n=1}^\infty E_n\right)=\lim\mu(E_n).$$ […]

About the Application of Cauchy-Schwarz (Basic): maximum of $3x+4y$ for $x^2+y^2 \leq 16$

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 […]

How find this minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n}a_{i}a_{i+1},a_{n+1}=a_{1}$

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 $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$, […]