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Probability inequality in Markov chain

$(EZ_n)^{2}=(EZ_n1_{Z_n >0})^{2}\leq EZ_n^{2}E1_{Z_n >0}^{2}=P(Z_n >0) EZ_n^{2}$. (I have used Cauchy-Schwarz-Bunyakovski inequality).
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0 votes

What are the minimum and maximum values here?

This is an immediate consequence of AM-GM. First, $u + 1/u \geq 2$ whenever $u > 0$. Lower bound: by inequality above, $\mu \geq {2^2}^2 = 16$, and this is attained with $\alpha = \beta = \gamma = ...
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1 vote

Contour Integral Involving $e^z$, a Semicircle, and Triangle Inequality

In THIS ANSWER, I showed that $|\cot(\pi z)|$ is bounded on the circle $|z|=(N+1/2)$ for $N\in \mathbb{N^+}$. Analogously, it is straightforward to show that $|\tanh(z)|$ is bounded on the circle $N\...
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0 votes

let $I_k=\int_0^\infty\frac{x^k}{p(x)}dx$ for which $k$ is $I_k$ smaller?

Perhaps, it is possible to reduce the comparison problem only to the comparison of $\int x^k$ by applying the product rule to get rid of the influence of $p(x)$ on $x^k$ under the integral sign. Then ...
0 votes

How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$?

For $N\geq 1$, both expressions are positive and less than 1. So we may compare their reciprocals; the one with smaller reciprocal is larger. For the first one, $${(\frac{2n+1}{2n+2})}^{-1}=\frac{2n+2}...
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3 votes

How to show that $\ln(2) > \frac{2}{3}$

Indeed, considering series like $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...\tag{1}$$ can be interesting but this one has a too slow rate of convergence. Instead use this one (derived from (1)): $$\ln\...
7 votes
Accepted

How to show that $\ln(2) > \frac{2}{3}$

$\ln(2) > \frac{2}{3}$ is equivalent to $e^2 < 8$, so we need an upper bound for $e$. A standard trick is to replace the “tail” of the exponential series by a geometric series, which can be ...
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5 votes

How to show that $\ln(2) > \frac{2}{3}$

You can define $\ln 2$ by $ \displaystyle \ln 2 = \int_1^2 \frac 1t \, dt$. Any Riemann sum for this integral with right endpoints will be less than $\ln 2$ because $\dfrac 1t$ is decreasing. In ...
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0 votes

Converse to Jensen's inequality for $1/x$ on a positive bounded interval?

We want to determine an upper bound for the function $$ h(x_1, \ldots, x_n) = \frac{1}{n^2} \sum_{k=1}^n x_k\sum_{k=1}^n \frac{1}{x_k} = \sum_{k, l=1}^n \frac{x_k}{x_l} $$ on the hypercube $[a, b]^n$....
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1 vote

generalization of Markov's inequality

Hint:- For positive numbers, is $a>b$ equivalent to $a^{p}>b^{p}$ ? If yes then what can you say about the event $\{|X|>c\}$ and $\{|X|^{p}>c^{p}\}$ ? . What can you then say about the ...
0 votes

Converse to Jensen's inequality for $1/x$ on a positive bounded interval?

$C = \frac{(a+b)^2}{4ab}$ is an upper bound for the ratio, and that bound is best possible. Proof: If $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ are real numbers in the interval $[m, M]$, $0 < m <...
  • 90.3k
0 votes

Converse to Jensen's inequality for $1/x$ on a positive bounded interval?

One can take $C(a, b) = b/a = 1 + (b-a)/a$. To see this, note that $$ \frac{\frac{1}{n}\sum_i x_i^{-1}}{((1/n)\sum_{i=1}^n x_i)^{-1}} = \frac{1}{n^2} \Big(\sum_{i} x_i\Big) \Big(\sum_i x_i^{-1}\Big) \...
  • 2,475
0 votes

Converse to Jensen's inequality for $1/x$ on a positive bounded interval?

Yes, there is a bound available. Let $t_i := \max \Big\{ \tfrac{(x_i - \overline{x})^2}{x_i^3}, \tfrac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}$. We have that the remainder is the average $\...
  • 2,475
0 votes

An urn initially contains one red and one blue ball question

(a) Notice that $P(X>k)$ is the probability of you geting a red ball from the first $k$ selections, that is $\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot \dfrac{3}{4}\dots \dfrac{k}{k+1} = \dfrac{1}{k+1}$, ...
  • 1,097
0 votes

Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$

I think your current description is not satisfactory. Take Case 3 for example. You said, This means, $f(x) \ge \frac {3a^2}{4}-5a+4$. Thus, for $f(x) > 0$, it is enough to take $\frac {3a^2}{4}-5a+...
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3 votes
Accepted

An upper bound of $\mathbb{P}(|S_n - \log n| \geq C \log n)$, where $S_n$ is a sum of $n$ independent Bernoulli$-\frac{1}{i}$ random variables

Using Chebycheff's Inequality (in reality we are just using Markov's inequality) $$P\bigg(\bigg|S_{n}-\ln(n)\bigg|\geq C\ln(n)\bigg)\leq \frac{E(|S_{n}-\ln(n)|^{2})}{C^{2}\ln^{2}(n)}.$$ Let us denote $...
2 votes

$n^3 \le (\sum_{i=1}^{n} x_{i}^{2})(\sum_{i=1}^{n} \frac{1}{x_i})^2$

A third proof using Cauchy-Schwarz inequality as you wish to see. $a)$ $ x_1^2+x_2^2+\cdots x_n^2 \ge \dfrac{(x_1+x_2+\cdots +x_n)^2}{n}$. $b)$ $\left(\left(x_1+x_2+\cdots +x_n\right)\left(\dfrac{1}{...
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4 votes

$n^3 \le (\sum_{i=1}^{n} x_{i}^{2})(\sum_{i=1}^{n} \frac{1}{x_i})^2$

A second proof using AM-GM inequality two times: $a)$ $x_1^2+x_2^2+\cdots + x_n^2 \ge n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}$. $b)$ $\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\cdots +\dfrac{1}{x_n}\right)^2 \ge \...
  • 4,308
3 votes
Accepted

$n^3 \le (\sum_{i=1}^{n} x_{i}^{2})(\sum_{i=1}^{n} \frac{1}{x_i})^2$

We have: $\text{LHS} =n^3 = \left(1+1+\cdots + 1\right)^3=\left(\displaystyle \sum_{i=1}^nx_i^{\frac{2}{3}}\cdot \frac{1}{x_i^{\frac{1}{3}}}\cdot \frac{1}{x_i^{\frac{1}{3}}}\right)^3\le \text{RHS}$ by ...
  • 4,308
4 votes
Accepted

Prove that $\int_0^\pi xa^{\sin x}dx\cdot\int_0^\pi a^{-\sin x}dx\geq\frac{\pi^3}{2}$ for $a > 0$

We have $$\int_0^\pi x a^{\sin x}\,\mathrm{d} x = \int_0^{\pi/2} x a^{\sin x}\,\mathrm{d} x + \int_{\pi/2}^\pi x a^{\sin x}\,\mathrm{d} x = \pi \int_0^{\pi/2} a^{\sin x}\,\mathrm{d} x$$ and $$\int_0^...
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3 votes

Logic behind equations and inequalities involvin absolute values

You can get the lower results from the upper ones. For example: $\begin{array}{c}&&& |x - a| & \leq & D \\ \implies & -D & \leq & x - a & \leq & D \\ \implies &...
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1 vote

Probability Question, sanity check

My understanding of your problem is, there are effectively two random variables representing squares $X$ and $X_n$. Suppose each random variable is the 'mass' of a square. $X + \epsilon$ is always a ...
2 votes
Accepted

Permutations with inequalities constraint - analytical solution

Apologies for prematurely closing your question as a duplicate. Although counting the things you want – linear extensions – for general posets is indeed hard, the special case of chained inequalities ...
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3 votes
Accepted

How to show that $\left| \frac{-x^2y-y^3-y}{x^2+y^2} \right| < \pi$?

Alright so we want to show that $$\left\lvert \frac{-x^2y-y^3-y}{x^2+y^2} \right\rvert < \pi$$ whenever $$1<\frac{1}{x^2+y^2}<4.$$ Now notice that then $$\left\lvert \frac{-x^2y-y^3-y}{x^2+y^...
  • 4,653
1 vote

How to show that $\left| \frac{-x^2y-y^3-y}{x^2+y^2} \right| < \pi$?

The problem with your solution is that by bounding separately the two expression and then putting them togheter we are not able to obtain a "good" estimation for the upper bound. For example ...
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1 vote
Accepted

Region of Convergence for the complex series $\sum {((z-1)/(z+2))^k} $

The ratio test is a sufficient but not necessary condition of convergence. For a geometric series with ratio $q$, the necessary and sufficient condition is $|q|<1$. Your complicated numerator is ...
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0 votes

If $ \frac{1}{4} \leq x^2 + y^2 \leq 1$, then is it true that $0 \leq \left|-(x^2+y^2)-1\right|\leq \frac{3}{4}$?

We have that $$\left|-(x^2+y^2)-1\right|=x^2+y^2+1>0$$ and $$\frac{1}{4} \leq x^2 + y^2 \leq 1 \iff \frac{5}{4} \leq x^2 + y^2 +1\leq 2$$ but $$0 \leq \left|-(x^2+y^2)-1\right|\leq \frac{3}{4} \iff ...
  • 142k
0 votes

If $ \frac{1}{4} \leq x^2 + y^2 \leq 1$, then is it true that $0 \leq \left|-(x^2+y^2)-1\right|\leq \frac{3}{4}$?

Inequality (1) is correct; you have reflected your given inequality across the origin. Inequality (2) is correct; you have shifted (1) in the negative direction. When working with inequalities, it is ...
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7 votes
Accepted

Maximum of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ for $a, b, c \ge 0$; $a+b+c=1$

We have, for all $x\in [0, 1]$, $$\frac{1}{18}x + \frac{1}{9} - \frac{1}{x^2 - 4x + 9} = \frac{x(x-1)^2}{18(x^2-4x+9)} \ge 0. \tag{1}$$ Using (1), we have $$\frac{1}{a^2 - 4a + 9} + \frac{1}{b^2 - 4b +...
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3 votes

Maximum of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ for $a, b, c \ge 0$; $a+b+c=1$

This is something of a brute force method. Let $f(a,b,c)=\sum_{a,b,c}1/ (a^2-4a+9)$ and $g(a,b,c)=a+b+c-1$ then calculate $\nabla f-\lambda \nabla g=0$ to find $$ {2a-4\over (a^2-4a+9)^2}-\lambda = 0 $...
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1 vote

Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$

I want to bring the most straightforward way to calculate this problem systematically. Also, you can check the attached graph for better representation. I write the quadratic equation $f_x(a)=4x^2-2ax+...
0 votes
Accepted

Showing that $\lfloor \sqrt{n^2+b}\rfloor=n \iff 0 \le b < 2n+1$

\begin{align} 0\leq b<2n+1 &\iff n^2 \leq n^2+b <n^2+2n+1\\ &\iff n^2\leq n^2+b <(n+1)^2 \\ &\iff n\leq \sqrt{n^2+b} <n+1 \\ &\iff \lfloor\sqrt{n^2+b} \rfloor = n \\ \end{...
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1 vote

Adding constants to the numerator and denominator of a fraction

No, $\frac{1}{2} \leq \frac{3}{4}$ and $\frac{4}{2} = \frac{2}{1}$, but $\frac{1+4}{2+2} = \frac{5}{4} > \frac{3+2}{4+1}=\frac{5}{5}$
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1 vote

Adding constants to the numerator and denominator of a fraction

This is not correct. For example: $$\frac{1}{2}=0.5 < 0.66\approx \frac{1}{1.5}.$$ If you add $\frac{5}{5}=\frac{1}{1}$ you get $$\frac{6}{7} \approx 0.85 > 0.8=\frac{2}{2.5}$$
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Adding constants to the numerator and denominator of a fraction

No: $\frac{2}{1}\leq\frac{1}{\frac 12}$ (in fact, this is an equality) but $\frac{2+2}{1+1}\not\leq\frac{1+1}{\frac 12+\frac 12}$.
  • 402
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Chiswell/Hodges Exercise 3.5.5 -- Inequality for the number of truth values in truth table

Let us discuss the calculation over a parsing tree example from the book. The formula $\phi$ is $(p_{1}\wedge (\neg(p_{0}\rightarrow p_{2})))$, depicted as: So, the number of nodes, $n$, is $6$. The ...
  • 1,130
6 votes

If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}<x<\frac{2014}{2015}$

We have $$\frac{1}{k}-\frac{1}{k+1} < \frac{1}{k^2} < \frac{1}{k-1} - \frac{1}{k}$$ and so (telescoping sum) $$\frac{1}{2} - \frac{1}{2016} < \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{1}{1} -...
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6 votes

If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}<x<\frac{2014}{2015}$

We have the bound $$\sum_{n=2}^{2015}\frac{1}{n^2}>\int_2^{2016}\frac{1}{x^2}dx=\frac{1}{2}-\frac{1}{2016}>\frac{1}{2}-\frac{1}{806}=\frac{402}{806}=\frac{201}{403}$$ by thinking about the graph ...
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2 votes
Accepted

prove that there exist $x_1\neq x_2, x_1,x_2\in [0,1]$ so that $\int_{x_1}^{x_2} f(x)dx = (x_1-x_2)^{2002}$

You're right that the IVT is useful here: Wlog we can assume $a < b$ and $c < d$, now define $$ g: [0, 1] \to \mathbb{R}\\ t \mapsto |F(ta+(1-t)c) - F(tb+(1-t)d)| - |t(a-b) + (1-t)(c-d)|^{2002} $...
2 votes

Prove that $(a-d)^2+(b-c)^2\geq 1.6$ if $a^2+4b^2=4$ and $cd=4$

Using $a^2 + 4b^2 = 4$ and $cd = 4$, we have \begin{align*} (a-d)^2 + (b-c)^2 &= (a-d)^2 + (b - c)^2 + \frac12 (a^2 + 4b^2 - 4)\\[5pt] &= \frac16(3a - 2d)^2 + \frac13(3b - c)^2 + \frac23c^2 +...
  • 25.9k
0 votes

Proving that $\sqrt{x^2+1}+x>0$ for all $x$

x = 0; the inequality is satisfied 1 > 0 x < 0; $\sqrt{x^2+1}-x > 0;$ x > 0; $\sqrt{x^2 + 1}+x > 0;$
1 vote

Let $x \in R$. Then, prove that $x^2+|x-6|>5$

Your attempt seems incorrect to me when you analyze the case $3$. Note that, if $x≤6$, then you have: $$ \begin{align} &x^2-x+6-5>0\wedge x≤6\\ \implies &\left(x-\frac 12\right)^2+\frac 34 &...
  • 8,774
2 votes
Accepted

Let $x \in R$. Then, prove that $x^2+|x-6|>5$

For your case $(3)$, notice that if $x<6$ then $|x-6|=6-x$. Then, your inequality becomes $$x^2+6-x>5$$ $$x^2-x+1>0$$ $$(x-\frac{1}{2})^2+\frac{3}{4}>0$$ which is true.
  • 1,276
6 votes

How do I make this 'intuitive' argument into a rigorous proof?

Your observation that $f(x)^3 ≤ f(x)$ when $0 ≤ f(x) ≤ 1$ is good. A similar observation extends this idea to the entire range: For all $y ≤ 1$, we have $y^3 ≤ y^3 + \frac14(1-y)(1 + 2y)^2 = \frac14 + ...
0 votes

Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$

I've revised your work as follows. The first step is fine $$ 0<x<2 \implies -\frac a2<2x-\frac a2<4-\frac a2$$ As a minor issue we should use $\iff$ instead of $\implies$. Case $-1:\,\,\,...
  • 142k
-2 votes

Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$

I've tried to answer in a way which skips over some of the tedious calculation so as to make it easier to understand what this question means. $$f(a)=4x^2-2ax+a^2-5a+4>0$$ Consider, for the moment, ...
  • 1,632
0 votes

Prove that $(a-d)^2+(b-c)^2\geq 1.6$ if $a^2+4b^2=4$ and $cd=4$

I came across this problem just recently and was compelled to obtain the result by resorting to the solver function in Excel. Indeed, we are given a system of the following four equations: $$\begin{...
  • 1
0 votes

Prove $2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2)$ for $x, y, z \ge 0$

Using $u^4 - (2u^3 - u^2) = (u-1)^2u^2 \ge 0$, we have $$x^4 + y^4 + z^4 \ge 2x^3 + 2y^3 + 2z^3 - x^2 - y^2 - z^2.$$ It suffices to prove that $$4(x^3 + y^3 + z^3) + 2xyz + 7 \ge 7(x^2 + y^2 + z^2).$$...
  • 25.9k
1 vote

Prove $\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1$ for $x,y,z > 0 $

By the Cauchy-Schwarz inequality, $$ (1+xy+xz)(1+y/x+z/x) \\ =\left(1,\sqrt{xy},\sqrt{xz}\right)^2 \left(1,\sqrt{ y/x},\sqrt{z/x}\right)^2\geq(1+y+z)^2 $$ which we can rearrange to $$ {1+xy+xz\over(1+...
0 votes

Prove that $x\sin A+y\sin B+z\sin C\leqslant \frac{\sqrt{\left( x^2+k \right) \left( y^2+k \right) \left( z^2+k \right)}}{k}$

Suppose $\cos B=t\in(-1,1)$ and we shall use the trick of the expansion of sin C in A, B and organizing it in the form of $A_1 \sin A+ A_2\cos A +A_3$ so that we can use the Cauchy inequality to turn ...

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