linear algebra – Inequality in the inverse numerical range of the core matrix

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}

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.