# Tag Info

0

I think that this statement is false: Take for example $e=4/3$, so the statement is $$x^2>-\frac{1}{4}\ \Rightarrow\ x<-\frac{1}{2}\ \text{or}\ x>\frac{1}{2}$$ $x=0$, for instance, is a counterexample.

0

ok so lets first consider the first inequality: $$\frac{x}{x+2}>0\tag{1}$$ for this to be true either $x>0$ so the top and bottom are both positive, or we can have $x<-2$ and so the solution for this inequality would be $x\in(-\infty,-2)\wedge(0,\infty)$. Now for the second: $$\frac{x+1}{x+2}<1\tag{2}$$ $$1-\frac{1}{x+2}<1$$ $$-\frac 1{x+2}&... 0 Hint:$$-(a+b)(a+c)(b+c)(a^2-ab-ac+b^2-bc+c^2)=abc(a+b+c)^2-(a^3+b^3+c^3)(ab+bc+ca)$$1 Your proof is correct. I don't see a simpler way than this to prove that. 1 Hint: Using CS inequality,$$(a^3+b^3+c^3)\left(\frac1a+\frac1b+\frac1c\right)\geqslant (a+b+c)^2$$0 Note that$$ac\ge bc \text{ } \text{ and } \text{ } c>0 \Rightarrow a\ge b$$and$$ac\ge bc \text{ } \text{ and } \text{ } c < 0 \Rightarrow a\le b$$Hence we have to consider two different cases. I will consider two cases where  \color{blue}{x-2>0} and \color{blue}{x-2<0}$$x(x-2) \geq 2(x-2) \text{ } \text{ and } \text{ } \color{blue}{x-2&...

0

In your last step you divide both sides of the inequality by $(x-2)$ which is negative for $x < 2$, and this would require reversing the inequality sign. Instead you could simply split your solution into two cases: $x\geq 2$ and $x< 2$, and derive the entire solution space by considering each case individually.

0

You couldn’t have divided both sides by $x-2$ As it can be negative. $$-1>-2 \not \Rightarrow 1>2$$

0

WLOG $b=a\sin2t,$ As $a>b>0,0<\sin2t<1$ WLOG $0<2t<\dfrac\pi2\implies 0<t<\dfrac\pi4 \ \ \ \ (1)$ $$\dfrac{-a+\sqrt{a^2-b^2}}b=\dfrac{-a+a\cos2t}{a\sin2t}=\dfrac{1-2\sin^2t-1}{2\sin t\cos t}=-\tan t$$ By $(1), -1<\tan t<0$

1

Let $f(x) = \log(x)$ on $(1,x)$, $a = 1$, $b = x$. Then invoke MVT to find $c \in (1,x)$ such that $1/c = f'(c) = (f(b)-f(a))/(b-a) = (\log x - \log 1)/(x-1)$. Use $c > 1$ to prove the desired inequality.

3

Use AM-GM inequality, $$\frac{x+y+z}{3} \ge \sqrt [3]{xyz}$$ $$x+y+z \ge 3$$ The minimum is $3$ and there's no maximum.

0

By geometry: The surface of equation $xyz=1$ (don't know its name) is a cubic with an "hyperbolic-like" shape, as any cross section by a plane of one constant coordinate is an hyperbola. It has a symmetry of order $3$ around the axis $x=y=z$, and is open towards infinity. The sections by the plane $x+y+z=c$ are closed curves, starting from $c=3$ ...

0

Since $a>b$, put $x=a/b >1$. Then the inequality you are trying to prove becomes $$\left|-x + \sqrt{x^2-1}\right| < 1.$$ Now this is quite straight forward. Infact, we can see $$\left|(-x + \sqrt{x^2-1})(x + \sqrt{x^2-1})\right| = 1,$$ and therefore $$\left|-x + \sqrt{x^2-1}\right| < \frac{1}{(x + \sqrt{x^2-1})}.$$

0

The quadratic equation $bx^2+2ax+b=0$ has two roots: $r=\frac{-a+\sqrt{a^2-b^2}}{b}$ and $s=-\frac{a+\sqrt{a^2-b^2}}{b}$. Obviously, $|s|>1$, so $|r|=\frac{1}{|s|}<1$ from Vieta's formulas.

1

Note that $0 < a^2 - b^2 < a^2$. So $\sqrt{a^2-b^2} < a$. Hence \begin{align} \left| \frac{-a + \sqrt{a^2-b^2}}{b} \right| &= \frac{a - \sqrt{a^2-b^2}}{b} \\ &= \dfrac{b}{a+\sqrt{a^2-b^2}} \\ &< 1 \end{align}

0

The number in question is the absolute value of the greatest root of $\frac{1}{2}b x^2+ax+\frac{1}{2}b$. Let $\alpha$ denote what is inside the absolute value. Then $\alpha$ must be negative since $a^2-b^2 < a^2$. Meanwhile, by symmetry, if $x_{0}$ is a non-zero root of this polynomial, then so is $1/x_{0}$. Since $\alpha$ is the greatest root and ...

4

Note that$$\left|-a+\sqrt{a^2-b^2}\right|=a-\sqrt{a^2-b^2},$$since $0<b<a$, and that\begin{align}a-\sqrt{a^2-b^2}<b&\iff a-b<\sqrt{a^2-b^2}=\sqrt{(a-b)(a+b)}\\&\iff(a-b)^2<(a-b)(a+b)\\&\iff a-b<a+b,\end{align}which is true, since $b>0$.

0

Alternative proof: Case 1: $a\ge b+c.$ Then $a\ge b, a\ge c$. $$\min(a,b+c)=b+c =\min(a,b)+ c=\min(a,b)+\min(a,c).$$ Case 2: $a \le b+c.$ Then $\exists x, y \ge 0, a=x+y$, such that $x\le b, y \le c$. (For example $x=\frac{ab}{b+c}, y=\frac{ac}{b+c}.)$ Then $$x \le a, x \le b \implies x \le \min(a,b)$$ $$y \le a, y \le c \implies y \le \min(a,c)$$ ...

4

It's an direct consequence of the Hermite-Hadamard inequality We have with $f(x)=px^{p-1}$ a convex function , $x\geq y\geq 0$ and $p\geq 1$: $$\frac{1}{x-y}\int_{y}^{x}f(x)dx\leq \frac{f(x)+f(y)}{2}$$ Or : $$\frac{x^p-y^p}{x-y}\leq \frac{px^{p-1}+py^{p-1}}{2}$$

1

When you inspect the two cases, you are saying that the condition described in the problem is the logical disjunction of the conditions coming from each case, not their logical conjunction. Therefore you then have to consider, as solution set, the union of their solution sets (which correponds to the points that satisfy at least one of the two conditions). ...

2

WLOG $x\ge y$ let $t=\frac{x}{y}\ge 1$ we have to prove $$f(t)=(p-1)t^p-pt^{p-1}+pt-p+1\ge 0$$ notice $$f'(t)=p(p-1)(t^{p-1}-t^{p-2})+p>0$$ thus $f(t)$ is increasing also $f(1)=0$ hance $f(t)\ge 0$ for all $x,y$ in domain given (because the case $x\le y$ can be proved analogously)

2

Almost you are reached out... Let $\epsilon>0$. Let $N=\frac{\epsilon}{2}$, Therefore for all $n>N$. $$\left| \frac{n^3+1}{n^4+1}-0 \right|= \left| \frac{n^3+1}{n^4+1}\right|\leq 2\frac{1}{n}<2\frac{1}{N}=\epsilon.$$

5

Wlog $x>y$. Then with $f(t):=x^p$, we have from the Mean Value Theorem that there exists some $\xi$ with $x>\xi>y$ such that $$\frac{x^p-y^p}{x-y}=\frac{f(x)-f(y)}{x-y}=f'(\xi) =p\xi^{p-1}\le p\max\{x^{p-1},y^{p-1}\}<p(x^{p-1}+y^{p-1}).$$

5

The issue is in the first line of simplification. $$a+\frac{a+b}{a}=a+1+\frac{b}{a}\neq 2+\frac{b}{a}.$$ If you make the correct simplification, you can continue $$\sqrt{\frac{\left(a+1+\frac ba\right)^2+\left(b+1+\frac ab\right)^2}{2}}\geq \frac{a+b+2+\frac ab+\frac ba}{2}=\frac{3+\frac ab+\frac ba}{2}\geq \frac 52,$$ and get the correct result.

2

For your case 3: $\min\{a,b\} + \min\{a,c\} = b+c$. Since $\min\{a,b+c\}\leq b+c$ (min is always less than both numbers) you get the inequality.

2

(Fill in the gaps. If you're stuck, show your work and explain why you're stuck.) Hint: Your quadratic is in $\cos x$, which is restricted to $-1 \leq \cos x \leq 1$. If you want to prove that a upward opening quadratic $f(x)$ on a restricted domain of $[p, q]$ is $< 0$, then we just need to show that $f(p) < 0$ and $f(q) < 0$. In particular, ...

0

Hint $$\cos x = 1 \implies 1+(1-a)-a^2<0 \implies 0 < a^2+a-2=(a-1)(a+2) \implies a<-2 \text{ or } a> 1.$$ Therefore $a$ can't be $-1$.

1

Here it is a proof based on probability and copulas theory for the sake of curiosity. Let $C:[0,1]^{2}\to[0,1]$ be a copula. Since it is nondecreasing in each coordinate and it has uniform margins, one has that \begin{align*} \begin{cases} C(u,v) \leq C(u,1) = u\\\\ C(u,v) \leq C(1,v) = v \end{cases} \Rightarrow C(u,v)\leq\min\{u,v\} \end{align*} Moreover, ...

5

You have $x-1\leq 0$ and $y-1\leq 0$ so $x+y-1$ is less or equal to both $x$ and $y$. Therefore $\min{(x,y)}\geq\max{(0,x+y-1)}$

0

This is just Cauchy Schwarz inequality, $\sum a_i^2 \sum b_i^2 \geq (\sum a_i b_i)^2$ Set $a_i = 2, b_i = t_i$, and we get that $$4n \sum t_i^2 \geq (\sum 2 t_i)^2.$$

1

Let $M := M(\epsilon,\Delta t) = \frac{\sqrt{\epsilon}}{\Delta t}\left(1 + \left(1 - \frac{1}{\epsilon}\right)\Delta t\right)$. Recall that for any random variable $X$ with a finite second moment and non-negative number $x$, Chebyshev's inequality states: $$\mathbb{P}(|X| > x) \leq \frac{\mathbb{E}[X^2]}{x^2}.$$ I'm guessing your application of the ...

1

How about this method? $$\frac{2(x + y)²}{2x² + y²}-3=\frac{2x^2+4xy+2y^2-(6x^2+3y^2)}{2x^2+y^2}=-\frac{4x^2+4xy+y^2}{2x^2+y^2}=-\frac{(2x-y)^2}{2x^2+y^2}\le0..$$

2

The inequality is not true. Take $x = 1, y = -1, n = 3$ then we have $2^3 \leq 3(1+1)$. It might help for you to state the original inequality.

0

The OP was not very clear if them were looking for a combinatorial interpretation, in such case: Consider the following function $Fix : S_n\longrightarrow P([n]),$ called the set of fixed points of a permutation, where $S_n$ is the set of permutations on $[n]$ and $P([n])$ is the powerset of $[n],$ defined as $$Fix(\pi)=\{i: \pi(i)= i \},$$ for some ...

0

Proof: Consider the function $c(t) = -\ln\sum_k p_k^t q_k^{1-t}$. First note the $c(t)$ is a concave function of $t$ that is positive in $[0,1]$. Moreover, it vanishes at $t = 0$ and at $t=1$. Therefore, the supremum of $c(t)$ occurs at some $t^* \in [0,1]$. The value of $c(t)$ at $t^*$ is the Chernoff distance: $$c(t^*) = C.$$ Similarly, the Bhattacharyya ...

1

If $n!>2^n$ multiply both sides by $(n+1)$ to get $$(n+1)!>(n+1)2^n> 2 \times 2^n > 2^{n+1}$$ with minor details to be filled.

0

multiply both sides by n+1 Then prove that $(n+1)!2^n$ is greater than $2^{n+1}$

0

I think it's simpler than you're suggesting. Are you familiar with proof by induction? Here's a hint: once you know that $4! > 2^4$, look at how you get from $4!$ to $5!$ and from $2^4$ to $2^5$. If you can prove it for $n = 5$ given that it's true for $n = 4$, can you continue that process?

1

The number $n!$ is the product of the $n-1$ numbers $2,3,\ldots,n$, each of which is greater than or equal to $2$ (and all of them but $2$ is actually greater than $2$) and, since $n>3$, at least one of them is greater than or equal to $4$. So, $n!>2^{n-2}\times4=2^n$

0

HINT I would proceed by Induction. The base case $n=2$ is already given, so let's inductively assume this holds for all $n$ up to $k-1$, and then prove it for $n=k$. Assume WLOG that $\lambda_0 \ne 0$ (otherwise, you must have some non-zero coefficient since they have to sum up to 1, so pick any non-zero coefficient and renumber them). Let $x = \sum_{i=0}^{k-... 1 Well$2x^2+y^2> 0$and$2(x+y)^2\ge 0$so So$\frac{2(x + y)^2}{2x^2+ y^2} \leq 3\iff2(x+y)^2 \leq3(2x^2 + y^2) \iff2x^2 + 4xy +2y^2 \leq 6x^2 + 3y^2 \iff0 \leq 4x^2 -4xy +y^2 \iff0 \leq(2x -y)^2 \iff$God's in his heaven and all's right in the world. 1 Let$y=l.x$; we have:$\frac{2[(l+1)x]^2}{x^2(l^2+2)}=\frac {2l^2+4l+2}{l62+2}=2+\frac{4l-2}{l^2+2}\frac{4l-2}{l^2+2}<1$because:$(l-2)^2>0l^2-4l+2+2>04l-2<l^2+2$If$l=2$the equality holds. 3 by C-S $$(2x^2+y^2)(1+2)\ge {(\sqrt{2}x+\sqrt{2}y)}^2=2{(x+y)}^2$$ 6 $$\frac{2(x + y)²}{2x² + y²} \le 3 \iff 2x^2+4xy+2y^2\le 6x^2+3y^2 \\\iff0 \le 4x^2-4xy+y^2=(2x-y)^2.\blacksquare$$ 0 Yuri Negometyanov gave $$\left(\frac{a+1}{a+b}\right)^a = \left(1 + \frac{b-1}{a+1}\right)^{-a} \ge 1 - a\cdot \frac{b-1}{a+1} = 2 - b + \frac{b-1}{a+1}. \tag{1}$$ This follows from the Bernoulli inequality$(1+x)^r \ge 1 + rx$for$x > -1$and$r \le 0$. Using (1), it suffices to prove that $$\frac{b-1}{1+a}+ \frac{c-1}{1+b}+ \frac{a-1}{1+c}\ge 0. \tag{2}... 1 I think I have an answer to this question but would appreciate any feedback! :) Firstly note that$$ P(N=0) \leq P(|N-E(N)| = E(N)) $$We know that$$ P(|N-E(N)| \geq E(N)) \leq \frac{Var(N)}{E(N)^2} $$using Chebyshev. We want to connect these two expressions. We can do this by noting that$$ P(|N-E(N)| \geq E(N)) = P(|N-E(N)| = E(N)) + P(|N-E(N)| > E(... 0 |a|-|b|≥-|b-a| since |b| always greater or equal to zero, there are only two cases: case 1: |a|-|b| = |a-b| this happens if a ≥ b case 2: |a|-|b| = |b-a| this happens if a ≤ b furthermore, |a-b|≥ 0 and |b-a| ≥ 0 therefore in both cases |a|-|b| ≥ -|b-a| rearrange again to prove |a|≥|b|-|b-a| 4 Equality holds for$x=1$, which suggests to use apply the mean value theorem to$f(x) - f(1)$with$f(x) = \sqrt x$. We get $$\sqrt x - \sqrt 1 = (x-1) \frac{1}{2\sqrt c}$$ for some$c$between$1$and$x$. Now consider the cases$0 \le x < 1$and$x > 1$and show that the right-hand side is always $$\le \frac{x-1}{2} \, .$$ (Note that we have ... 1 For$x \ge 0$we have: $$\sqrt x \leq \frac{x+1}{2} \iff x \le \frac{1}{4}(x+1)^2 \iff 4x \le x^2+2x+1 \iff 0 \le x^2-2x+1 \iff 0 \le (x-1)^2.$$ 1 It is not necessary to use MVT. $$0\le(\sqrt x-1)^2=x-2\sqrt x+1 \iff 2\sqrt{x}\le x+1 \iff \sqrt{x}\le\frac{x+1}{2}$$ I use$x\ge0$twice: first to be$\sqrt{x}$well defined. Second,$(\sqrt{x})^2=|x|=x\$.

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