### inequality - Plotting inequalities domain comb-style with JSXGraph

I want to draw axis line with comb-styled selection of inequality domain, like in this picture:Is it possible with JSXGraph?...Read more

I want to draw axis line with comb-styled selection of inequality domain, like in this picture:Is it possible with JSXGraph?...Read more

$a,b,c$ are positive reals with $abc = 1$. Prove that $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$I try to use AM $\ge$ HM.$$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{a^3(b+c)+b^3(a+c)+c^3(a+b)}$$Then how I proceed....Read more

Motivation of this question:This question is related to the expected stopping time of a stochastic process under two hypotheses. Especially, it answers the question "how many more samples are required such that a sequential test stops when there is a model missmatch compared to the case when there is no missmatch". I found that the ratio is ${D(f_0,f_1)}/{D(g_0,g_1)}$ if the null hypothesis is correct and ${D(f_1,f_0)}/{D(g_1,g_0)}$ if the alternative hypothesis is correct. I know that both should be greater than $1$ because $g_0$ and $g_1$ are...Read more

Let $x,y,z$ be positive real number satisfy $x+y+z=3$Find the maximum value of $P=\frac{2}{3+xy+yz+zx}+(\frac{xyz}{(x+1)(y+1)(z+1)})^\frac{1}{3}$...Read more

Consider a right angles triangle. Let, c be it's hypotenuse and a,b it's other sides.Then prove,$ \sqrt { \frac {s(c-a-b)}{2∆} } $ is complex.Where ∆= area of the triangle and s is semi perimeter.I am not able to tackle this questions second part which is the above fraction can be proven equal to i.Proving it is complex is easy by the triangle equality. But how to prove that it is equal to i. Also, is there any other way to prove that the above fraction is complex.Edit- Including the other method, there are basically 2 ways for proving it is ...Read more

Suppose that I have three probability density functions $f(t)$, $g(t)$ and $h(t)$ (cumulative density distributions are $F(t)$, $G(t)$ and $H(t)$) respectively, for which the following inequalities are valid:$\int_{-\infty}^{\infty} f(t)\cdot G(t)dt \geq \int_{-\infty}^{\infty} g(t)\cdot F(t)dt$$\int_{-\infty}^{\infty} g(t)\cdot H(t)dt \geq \int_{-\infty}^{\infty} h(t)\cdot G(t)dt$$\int_{-\infty}^{\infty} f(t)\cdot F(t)dt = \int_{-\infty}^{\infty} g(t)\cdot G(t)dt = \int_{-\infty}^{\infty} h(t)\cdot H(t)dt = 1/2$$\int_{-\infty}^{\infty} f(t)\c...Read more

I have the following sets$$A = \{x \in [0, 1]^4 \mid 83x_1 + 61x_2 + 49x_3 + 20x_4 \leq 100\}$$$$B = \{x \in [0, 1]^4 \mid 4x_1 + 3x_2 + 2x_3 + 1x_4 \leq 4\}$$How can I prove that $B \subset A$?...Read more

$x$ and $y$ are real numbers such that $ x\ge 0$ and $y \ge 0$.If $$ x + y \le 5$$$$ x + 2y \ge 8$$Then what is the minimum value of $5x+y$?There is something wrong in my approach.I write the first inequality as $$ 5 \ge x + y$$Adding with the second inequality yields, $$5 + x + 2y \ge x + y + 8$$Or $$ y \ge 3$$ Thus $$ 9y \ge 27$$Multiplying the second inequality by $5$ we obtain $$ 5x + 10y \ge 40$$ $$9y \ge 27$$Hence $5x+y \ge 13$...Read more

I have a problem as follows. Really appreciate if anyone can give me some suggestions.I have $4000$ binary variables $\{x_0, x_1,...x_{3999}\}$ and $4000$ inequalities which have both binary addition (denoted as $+'$) and normal (arithmetic) addition (denoted as $+$). For example, inequality $1$: $$(x_0 +' x_{39} +' x_{71} +' x_{3191}) + (x_{1} +' x_{44} +' x_{182} +' x_{2142}) + ... \leq 1$$inequality $2$: $$(x_3 +' x_9 +' x_{39}) + (x_{1} +' x_{90}) + ... \leq 1$$...inequality $4000$: $$(x_{99} +' x_{51} +' x_{1191}) + (x_{171} +' x_{1441} +'...Read more

Let $O(0,0), A(3,0), B(3,2)$ and $C(0,2)$ be four vertices of a rectangle. Let$$d(P,OA)≤\min {\Bigl(d(P,AB),d(P,BC),d(P,OC)\Bigl)}$$where $d$ denotes the distance of the point $P$ with the line segments given. Let $S$ denote the region consisting of all those points $P$ inside the rectangle $OABC$ which satisfies the above inequality. Find the area of region $S$.The last time I asked this question(which was almost a week ago), this problem was put on hold and then closed. I am preparing for an entrance exam and I came across this problem in my ...Read more

The binomial expansion of $(1+n)^k$ is$$(1+n)^k=1+\binom{k}{1}n+\binom{k}{2}n^2+\cdots+\binom{k}{k}n^k.$$If $n=1$, then the term in the middle is the largest, i.e. when $i=\lfloor k/2\rfloor$ and $i=\lceil k/2\rceil$.What about for other integers $n$? Which term is the largest?...Read more

So I'm trying to prove that for $\frac{1}{2}< x \leq 1$ we have$$\sum_{\lceil nx \rceil}^{n}{n \choose k} \leq 2^{nh(x)}$$I've managed to prove that $$\sum_{0}^{\lfloor nx \rfloor}{ n\choose k}\leq2^{nH(1-x,x)}$$But when I try to progress from the fact that $\sum_{0}^{n}{n \choose k}=2^{n}$ I seem to be getting the inequality sign the wrong way round. Can anyone help?...Read more

While reading an article on Hoeffding's Inequality, I came across a curious inequality. Namely$$\cosh x \leq e^{x^2/2} \quad \forall x \in \mathbb{R}$$I tried many ways to prove it and finally, the Taylor series approach worked:$$e^x = 1 + x + \frac{x^2}{2!} + \cdots$$$$e^{-x} = 1 - x + \frac{x^2}{2!} - \cdots$$ Adding the two and dividing by 2 (This operation being justified as both series converge),we get$$\cosh x = 1 + \frac{x^2}{2!}+ \frac{x^4}{4!} + \cdots$$ Expanding $e^{x^2/2}$ yields$$e^{x^2/2} = 1 + \frac{x^2}{2!}+ \frac{x^4}{4\times...Read more

My question is strictly related to this question.Frechet Differentiability versus Strict DifferentiabilityThe author said it can be proved that $f(x) - f(y) \le 3x(x - y)$ for all $x > y > 0$. After several attempts, I have not found how this could be proved.I did find out how the author built his function though. The piece-wise linear function $f$ was built from $x^2$ by using the points on the graph at $x = 2^j$ and $x = -2^j$, $j \in \mathbb{Z}$ and connect them all together by the increasing order of $j$.Even with this discovery, I ca...Read more

Let $d$ be a positive integer which is not the square of any integer, $x,y \in \mathbb Z$ and $u:=x+y \sqrt d$$u$ is such that $$ u \geq 1 \;\text{and} \; |u \overline u|=|(x+y \sqrt d)(x-y \sqrt d)|=|x^2-dy^2|=1 $$Prove that $$x\geq 1 \; \text{and} \; y \geq 0$$So far, I've tried case working (whether $x^2-dy^2=1$ or $x^2-dy^2=-1$)Thanks for your help, especially if you got a more straightforward approach....Read more