I have to prove that for $sin(0,1)$, $uinmathcal{S}(mathbb{R}^n)$, ($u$ is a Schwartz’s function):

$$ |(-Delta)^su(x)|leq c_{n,s}|x|^{-n-2s},quadforall xinmathbb{R}^nsetminus B_1(0), $$

for some $c_{n,s}>0$, where:

$$(-Delta)^su(x):=-frac{C(n,s)}{2}int_{mathbb{R}^n}frac{u(x+y)+u(x-y)-2u(x)}

{|y|^{n+2s}},dy,quadforall xinmathbb{R}^n, $$

is the fractional laplacian. I have no idea. Any help would be appreciated.

# Tag: inequality

## triangle inequality – Can any norm be used in a product metric

Given a norm $|cdot|:Bbb{R}^2 to Bbb{R}$ and metric spaces $(X,d_X), (Y,d_Y)$, we define $D((x_1,y_1),(x_2,y_2)) = |(d_X(x_1,x_2),d_Y(y_1,y_2))|$. Is $D$ always a metric? I know it is when working with the 1-norm, 2-norm and $infty$-norm.

Currently, I am aware that the norms on $Bbb{R}^2$ are strongly equivalent, in the sense that for any two norms $|cdot|,||cdot||$, there exists positive reals $c_1,c_2$ such that $c_1|v|le ||v||le c_2|v|$. However, if $(X,d)$ is a discrete metric space, there exist functions that are strongly equivalent to $d$ but violate the triangle inequality, so I’m not sure how useful norms being equivalent are.

Is there some trick I am missing which can prove that $D$ always satisfices the triangle inequality, or is there a counterexample to my claim?

## real analysis – Prove inequality in metric spaces

We prove that the statement: Let $(mathbb{R}^n,d)$ is a

euclidean space, $X$ is a non-empty closed subset of $mathbb{R^n}$. If $xin mathbb{R}^n-X$, then exist $yin X$ such that for all $zin X$, $d(x,y)leq d(x,z)$.

I thought going through a contradiction, it is not easy for me.

On the other hand, i tried to find the existence, but

i failed it. I am interested in this problem, can you give me a hint?

## inequality – $sum frac {1} {alpha + log_a {b}} le frac {2}{alpha}$

This is another olympiad-problem my teacher gave us to train.

$$sum frac {1}{alpha + log_a{b}} le frac {2}{alpha}$$ with $alpha in (0, 2)$ and $a, b, c in (0, 1) cup (1, infty)$. After working a little bit with it, we get $$sum log_{a ^ alpha b} {a} le frac {2}{alpha}$$, but I don’t know how to go on.

## trigo inequality prove for all x between -pi to pi

$$

forall xinleft(-pi,piright):sin(pi-x)leqpi-x

$$

Hi can you please help me with this?

## inequality – Is the following property true of any repeating cycle of $n$ real values?

I have found it challenging to state this inductive argument about a cycle of reals clearly and concisely.

I would greatly appreciate if someone could show me how to make the same argument in a more standard way or offer suggestions for tightening up the argument.

**Let**:

- $%$ be the modulo operation
- $c_1, c_2, dots, c_n$ form a repeating cycle of $n$ reals with:

$$c_{i+n} = c_i$$ - $M(c_1, c_2, dots, c_n) = sumlimits_{i=1}^n c_i$
- $f_{k,n}(c_i) = c_{(i+k) % n}$

**Claim:**

There exists $i$ such that for all $j ge 0$:

$$sumlimits_{t=0}^jf_{t,n}(c_i) le dfrac{(j+1)M(c_1,dots,c_n)}{n}$$

**Argument:**

(1) Base Case: $n=2$

- This follows since min$(c_1,c_2) le dfrac{c_1+c_2}{2}$

(2) Assume that it is true for any such cycle consisting of up to $n ge 2$ reals.

(3) Inductive Case: Let $c_1, c_2, dots, c_{n+1}$ form a cycle of $n+1$ reals with:

$$c_{i+n+1} = c_i$$

(4) There exists $k$ where $c_k le dfrac{M(c_1,dots,c_{n+1})}{n+1}$

(5) Define $d_1, d_2, dots, d_n$ such that:

$$d_i = begin{cases}

c_i, & i < k\

c_{i+1} & i ge k\

end{cases}$$

so that:

$$frac{M(d_1, dots, d_n)}{n} le frac{M(c_1, dots, c_{n+1})}{n+1}$$

(6) By assumption, there exists $i$ such that for all $j ge 0$:

$$sumlimits_{t=0}^j f_{t,n}(d_i) le frac{(j+1)M(d_1, dots, d_n)}{n}$$

(7) Case 1: $i < k$

$$sumlimits_{t=0}^j f_{t,n}(c_i) = sumlimits_{t=0}^j f_{t,n}(d_i) le frac{(j+1)M(d_1, dots, d_n)}{n} le frac{(j+1)M(c_1, dots, c_{n+1})}{n+1}$$

$$sumlimits_{t=0}^j f_{t,n}(c_i) = c_k + sumlimits_{t=0}^{j-1} f_{t,n}(d_i) le frac{M(c_1,dots,c_{n+1})}{n+1} + frac{(j)M(d_1, dots, d_n)}{n} le frac{(j+1)M(c_1,dots,c_{n+1})}{n+1}$$

(8) Case 2: $ige k$

$$c_i le frac{M(c_1, dots, c_{n+1})}{n+1}$$

$$sumlimits_{t=0}^j f_{t,n}(c_i) = c_k + sumlimits_{t=0}^{j-1} f_{t,n}(d_i) le frac{M(c_1,dots,c_{n+1})}{n+1} + frac{(j)M(d_1, dots, d_n)}{n} le frac{(j+1)M(c_1,dots,c_{n+1})}{n+1}$$

## Obtaining an inequality for cauchy criterion for series.

I have the following sum but I cannot bound it nicely, here is what l have tried but I’m not sure if it is correct. Here $m>n$

$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}$

$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}<frac{1}{(n+1)

!}sum_{k=n+1}^{m}(k+2)$

$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}<frac{m^2-nm+2m-2n}{(n+1)!}$

$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}<frac{(m-n)(m+n+2)}{(n+1)!}$

$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}<frac{(3m^2)}{(n+1)(n)}$

$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}<frac{(3m^2)}{frac{m+2}{2}(frac{m}{2})}$

## Explanation to an unmotivated step in an inequality.

Follows the original problem:

Let $a, b, c$ be non-negative real numbers. Prove that

$$

frac{a^2}{a^2 + 2left(a + bright)^2} +

frac{b^2}{b^2 + 2left(b + cright)^2} +

frac{c^2}{c^2 + 2left(c + aright)^2} geqslant

frac{1}{3}

$$

**This is my approach:**

The inequality is equivalent to

$$

frac{1}{1 + 2left(1 + frac{b}{a}right)^2} +

frac{1}{1 + 2left(1 + frac{c}{b}right)^2} +

frac{1}{1 + 2left(1 + frac{a}{c}right)^2} geqslant

frac{1}{3}

$$

Hence, Let

$x$ be $displaystyle frac{b}{a}$,

$y$ be $displaystyle frac{c}{b}$ and

$z$ be $displaystyle frac{a}{c}$.

Now the inequality is

$$

frac{1}{1 + 2left(1 + xright)^2} +

frac{1}{1 + 2left(1 + yright)^2} +

frac{1}{1 + 2left(1 + zright)^2} geqslant

frac{1}{3} quadtextrm{with}quad xyz = 1

$$

$$

frac{xyz}{xyz + 2(x + xyz)^2} +

frac{xyz}{xyz + 2(y + xyz)^2} +

frac{xyz}{xyz + 2(z + xyz)^2} geqslant

frac{1}{3}

$$

$$

frac{xyz}{xyz + 2x^2(1 + yz)^2} +

frac{xyz}{xyz + 2y^2(1 + zx)^2} +

frac{xyz}{xyz + 2z^2(1 + xy)^2} geqslant

frac{1}{3}

$$

$$

frac{yz}{yz + 2x(1 + yz)^2} +

frac{zx}{zx + 2y(1 + zx)^2} +

frac{xy}{xy + 2z(1 + xy)^2} geqslant

frac{1}{3}

$$

$$

frac{y^2z^2}{y^2z^2 + 2(1 + yz)^2} +

frac{z^2x^2}{z^2x^2 + 2(1 + zx)^2} +

frac{x^2y^2}{x^2y^2 + 2(1 + xy)^2} geqslant

frac{1}{3}

$$

By $T_2$‘s Lemma,

$$

frac

{left(xy + yz + zxright)^2}

{x^2y^2 + y^2z^2 + z^2x^2 +

2left((1 + yz)^2 + (1 + zx)^2 + (1 + xy)^2right)} geqslant

frac{1}{3}

$$

Hence, after simplification,

$$

3(x + y + z) geqslant

2(xy + yz + zx) + 3

$$

Substitute back $a,b,c$.

$$

3left(

frac {a} {b} +

frac {b} {c} +

frac {c} {a}

right) geqslant

2left(

frac{a}{c} +

frac{b}{a} +

frac{c}{b}

right) + 3

$$

I think that it looks easy, but if someone has a way, it might become a solution.

**Book’s Approach:**

It is the same till the second inequality in my approach, but then without any motivation goes on to assume $$x = frac{np}{m^2}, y = frac{mp}{n^2}, z = frac{mn}{p^2}.$$ And then it is just $T_2$‘s Lemma and some simplification to get squares on the greater side.

I don’t see anything natural in assuming that and neither there is a motivation to take such step, besides it gives proof.

But my question is how does the author assume such a thing? And is there a way to complete my proof?

Thanks!

## real analysis – Inequality involving sigmoid function

Let $sigma$ denote the sigmoid function $sigma(x) = frac{1}{1+e^{-x}}$, let $x,y in mathbb{R}$. Given the following two conditions: $|sigma(-x) – sigma(y)| < epsilon$ and $x – y > c > 0,$ where $epsilon$ can be regarded as a small positive number and $c$ as a large positive number. Can we draw the conclusion that $x > frac{c}{2} – f(epsilon), y <-frac{c}{2} + f(epsilon)$, where $f(epsilon)$ is some function of $epsilon$.

## assumptions – Why a complementary inequality for function values does not simplify to False?

I have a general assumption of a function to be positive. The direct inequality (f(x)>0) does simplify to True, but a complementary one (f(x) < 0) does not simplify at all despite the fact that I explcitly assumed that funciton is not zero.

```
$Assumptions = K(_) > 0 && K(_) != 0
FullSimplify(K(15) > 0)
True
FullSimplify(K(12) == 0)
False
FullSimplify(K(15) < 0)
K(15) < 0
```

Thansk in advance,