Leave $ k (.,.) $ be a function that takes two vectors as input and generates a scalar as follows

begin {align}

mathcal {k} (x, y) = exp (- frac {|| x-y || _2 ^ 2} {2})

end {align}

where $ || x || _2 $ denotes the $ 2- $rule. Now leave $ x_1, dots, x_m $ be $ m $ vectors in $ mathbb {R} ^ d $. Let me define the $ m times m $ matrix $ mathbf {K} $ so that your tickets are given as

begin {align}

mathbf {K} _ {ij} = mathcal {k} (x_i, x_j)

end {align}

For any vector $ x $let me define the $ m times 1 $ vector $ mathcal {K} _x $ as

begin {align}

mathcal {K} _x = begin {bmatrix}

mathcal {k} (x, x_1) \ vdots \ mathcal {k} (x, x_m)

end {bmatrix}

end {align}

Now I'm curious if the inequality tracking is valid for everyone. $ x in mathbb {R} ^ d $

begin {align}

mathcal {K} _x ^ T mathbf {K} ^ {- 1} mathcal {K} _x leq 1

end {align}

# Tag: inequality

## plotted: 3D graph with an inequality condition for parameter values

I'm trying to `ListPlot3D`

the next function

$ f = frac {- frac {(t-1) (ds) left (2 cd (d (q-1) + s) -s left (-2 d ^ 2 + d (s + 2) + s ^ 2 right) right)} {2 s ^ 2} + d ^ 2-d sd t + st} {s} $

against $ c in [0,1]$ Y $ q in [1,2]$ under the conditions of $ s = 2 $, $ d = 0.8 $, $ t = 0 $, $ 0 leq c leq 1 $Y $ 1 leq q leq frac {1} {c} $.

I'm struggling with how to reflect the last condition, that is, $ 1 leq q leq frac {1} {c} $ in my Mathematica code. used to `Assumption`

But it does not work

Here is the code that I tried:

`To block[{s=2d=08t=0}f=(d^2-ds-((ds)(2cd(d(-1+q)+s)-s(-2d^2+s^2+d(2+s)))(-1+t))/(2s^2)-dt+st)/s;maxn=Flat[Table[{cqf}{c011}{q121}Assumptions->{0<c<=11[{s=2d=08t=0}f=(d^2-ds-((d-s)(2cd(d(-1+q)+s)-s(-2d^2+s^2+d(2+s)))(-1+t))/(2s^2)-dt+st)/s;maxn=Aplanar[Mesa[{cqf}{c011}{q121}Suposiciones->{0`{0 <c <= 11 <= q<= 1/c}], 1];] {ListPlot3D[maxn, AxesLabel -> {"c", "q", "V"}]}

## inequality: given three non-negative numbers $ x, y, z $, find the range of $ k = constant $ for the following to be true.

Given three non-negative numbers $ x, y, z $, find the range of $ k = constant $ so the following is true:

$$ sum limits_ {cyc} x (1+ xy- y) (yz- z + 1) – 3xyz- kx (x- y) (y- z) geqq 0 $$

** observation.** For three non-negative numbers. $ x, y, z $, for $ k = 1 $ Y $ y equiv { rm mid} {x, y, z } $, we can also try

$$ sum limits_ {cyc} x (1+ xy- y) (yz- z + 1) – 3xyz geqq 0 $$

## Real analysis – Very strong inequality in the style of C.p Niculescu

I am interested in the following problem:

Leave $ a, b, c, d> 0 $ such that $ a geq b geq c geq d $ then we have:

$$ | a + d- sqrt {ad} – Big (a ^ ad ^ d Big) ^ { frac {1} {a + d}} | geq | b + c- sqrt {bc} – Big (b ^ bc ^ c Big) ^ { frac {1} {b + c}} | $$

In fact, it is a new refinement of my last inequality New limit for Am-Gm of 2 variables

It is very strong and I try to use the inequality of niculescu:

Leave $ f (x) $ Being a function twice differentiable and convex in $ I $ then for $ a geq b geq c geq d $ Y $ a, b, c, d in I $The inequality of Niculescu states that:

$$ f (a) + f (d) -f Big ( frac {a + d} {2} Big) geq f (b) + f (c) -f Big ( frac {b + c} {2} Big) $$

But I think it's too weak and I'm really lost. I recognize that I'm just an amateur and creating problems is easy, but solving it is difficult.

If you have a clue it would be good.

Thanks again and again!

## inequality: show that $ (abc + xyz) left ( frac1 {ay} + frac1 {bz} + frac {cx} right) geq3 $

Leave $ a, b, c, x, y, z in mathbb {R} _ + $ such that $ a + x = b + y = c + z = 1. $ Test inequality

$$ (abc + xyz) left ( frac1 {ay} + frac1 {bz} + frac1 {cx} right) geq3 $$

I tried using AM-HM to get

$$ (abc + xyz) left ( frac1 {ay} + frac1 {bz} + frac1 {cx} right) geq9 frac {abc + xyz} {ay + bz + cx} $$

and he wrote it as $$ frac {abc + xyz} {ay + bz + cx} = frac {1- (a + b + c) + (ab + bc + ca)} {(a + b + c) – (ab + bc + ca)} = frac1 {a + b + c-ab-bc-ca} -1 $$I mean, I have to prove that

$$ a + b + c-ab-bc-ca leq frac34 $$

I tried using Cauchy-Schwarz after this as $$ (a (1-b) + b (1-c) + c (1-a)) leq sqrt {(a ^ 2 + b ^ 2 + c ^ 2) ((1-a) ^ 2 + (1-b) ^ 2 + (1-c) ^ 2)} $$

But I had no idea to simplify it further.

Any help would be appreciated!

## [ Newborn & Baby ] Open question: Name one thing that defends the GOP that is not based on inequality?

[ Newborn & Baby ] Open question: Name one thing that defends the GOP that is not based on inequality? .

## Real analysis – Holder of the inequality with respect to the convex function.

Dice $ a, T> 0 $, for the inequality of Holder we have

$$

int_T ^ {T + a} f (s) ds leq left ( int_0 ^ {T + a} | f (s) | ^ 2ds right) ^ {1/2} cdot sqrt {a} .

$$

We have similar results if we replace $ | x | ^ 2 $ By some convex function? I mean, let's go $ Psi $ be a convex function such that $ Psi (x) / | x | a infty $ as $ | x | a infty $, we have

$$

int_T ^ {T + a} f (s) ds leq int_0 ^ T Psi (f (s)) ds cdot Phi (a),

$$

where $ Phi (a) $ it goes to 0 like $ a to0 $?

## Inequality – $ 2 left[(mz)^2 + (px)^2 + pmy^2right] + left[(nz)^2 + (nx)^2 + (py)^2 + (my)^2right] ge 2 (2pmzx + pnzy + mnxy + pnxy + mnzy) $

$ x, y, z $ Y $ m, n, p $ are real Try / disprove that $$ 2 large left[(mz)^2 + (px)^2 + pmy^2right] + left[(nz)^2 + (nx)^2 + (py)^2 + (my)^2right] ge 2 (2pmzx + pnzy + mnxy + pnxy + mnzy) $$

That is the problem and I do not know how to do it.

My original idea is to convert the difference between the left and right sides into a sum of squares and then use the Cauchy-Schwarz inequality (it was not accredited to Bunyakovsky) but it was not successful.

## postfix: Why does LDAP not support inequality lookups in shadowExpire?

When I try to search on my OpenLDAP 2.4.42 server `(shadowExpire <= 18074)`

, I do not get results However, a search for `(shadowExpire = 12671)`

Several results are obtained.

My ultimate goal is to periodically regenerate a Postfix map for `check_recipient_access`

to reject the mail to the expired accounts.

Looking at the server `/etc/ldap/schema/nis.schema`

, I see:

```
attributetype (1.3.6.1.1.1.1.10 NAME & # 39; shadowExpire & # 39;
EQUALITY integerMatch
SYNTAX 1.3.6.1.4.1.1466.115.121.1.27 UNIQUE VALUE)
```

This is in accordance with RFC 2307 Section 3, which specifies:

```
(nisSchema.1.10 NAME & # 39; shadowExpire & # 39;
EQUALITY integerMatch
SYNTAX & # 39; INTEGER & # 39; ONLY ONE VALUE)
```

Both suggest that `ShadowExperience`

it only supports equality searches only.

However, RFC 2307bis has:

```
(1.3.6.1.1.1.1.10 NAME & # 39; shadowExpire & # 39;
EQUALITY integerMatch
ORDERS integerOrderingMatch
SYNTAX 1.3.6.1.4.1.1466.115.121.1.27
SIMPLE VALUE)
```

… what's wrong with it `ORDERS integerOrderingMatch`

, allowing the inequality search to work.

Can I hack my server? `nis.schema`

to include `ORDERS integerOrderingMatch`

? The underlying representation would still be a `WHOLE`

, so it must be harmless, right? (Despite the warning in the OpenLDAP Administrator's Guide that says "you should not modify any of the schema elements defined in the provided files.)

## How to solve this inequality?

How to solve this inequality?

log x + 2 log (x-2) -6 log x log (x-2) <0

I have tried several methods but I can not solve this inequality. PLEASE HELP.