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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.

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probability: why is the event $ {X_ {(j)} le x_i } $ equivalent to the event $ {Y_i ge j } $?

From the statistical inference of Casella and Berger:

Leave $ X_1, points X_n $ Be a random sample of a discrete distribution.
with $ f_X (x_i) = p_i $, where $ x_1 lt x_2 lt dots $ they are possible
values ​​of $ X $ in ascending order. Leave $ X _ {(1)}, dots, X _ {(n)} $
denotes the order statistics of the sample. Define $ Y_i $ as the number of $ X_j $ that are less than or equal to
$ x_i $. Leave $ P_0 = 0, P_1 = p_1, dots, P_i = p_1 + p_2 + dots + p_i $.

Yes $ {X_j le x_i } $ it's a "success" and $ {X_j gt x_i } $ it is a "failure", then $ Y_i $ it is binomial with parameters $ (n, P_i) $.

Then the event $ {X _ {(j)} le x_i } $ is equivalent to the event $ {Y_i ge j } $

Can anyone explain why these two are equivalent?

$ {X_ {(j)} le x_i } = {s in text {sun} (X _ {(j)}): X _ {(j)} (s) le x_i } $

$ {Y_i ge j } = {s & # 39; in text {sun} (Y_i): Y_i (s & # 39;) ge j } $

I have trouble understanding how these functions of random variables show this equivalence.

Show the inequality $ sec x ge 1+ (x ^ 2/2) $

Test inequality

$ sec (x) ge 1+ (x ^ 2/2) $

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partial derivative – Ge the canonical form of the equation

Dice $$ x ^ 2u_ {xx} + 2xyu_ {xy} – 3y ^ 2u_ {yy} – 2xu_x + 4yu_y + 16x ^ 4u = 0 $$
I have to find the canonical form. The determinant is the following $$ D = 16x ^ 2y ^ 2 $$
I considered cases when $ x $ Y $ and $ can be $ 0 $. When calculating for the remaining case (where both $ x $ Y $ and $ It is not $ 0 $ and the type is hyperbolic), I'm not getting its canonical form. I'm receiving this
$$ (- 3y ^ 2 – 6x ^ 3y + 9x ^ 6) u _ { xi xi} + (-3y ^ 2 + frac {2y} {x} + frac {1} {x ^ 2} ) u _ {eta} + (-6y ^ 2 – 6x ^ 2 + frac {2y (1-3x ^ 4)} x) u _ { eta xi} + ( frac {-4} {x} + 4y) u_ eta + 4yu_ xi + 16x ^ 4u = 0 $$
But the point is that this does not meet the requirements for the canonical form of hyperbolic type.
Any help would be appreciated.

precalculus algebra: show the inequality $ sum x + 6 ge 2 ( sum sqrt {xy}) $

Leave $ x; Y; z in R ^ + $ such that $ x + y + z + 2 = xyz $. Such that $$ x + y + z + 6 ge 2 ( sqrt {xy} + sqrt {yz} + sqrt {xz}) $$


This inequality is not homogeneous and look at the condition that I thought
I would replace the variables $ x; Y; z $ such that

+)$ x ^ 2 + y ^ 2 + z ^ 2 + 2xyz = 1 $. Leave $ x = frac {2a} { sqrt { left (a + b right) left (a + c right)}} $

+)$ xy + yz + xz + xyz = 4 $. Leave $ a = frac {2 sqrt {xy}} { sqrt { left (y + z right) left (x + z right)}} $.Leave $ x = frac {2a} {b + c} $

but failed. Please explain to me how I can obtain this substitution (if I have a solution by substitution)

I also tried to solve it by $ u, v, w $.Leave $ sum_ {cyc} x = 3u; sum_ {cyc} xy; Pi_ {cyc} a = w ^ 3 (3u + 2 = w ^ 3; u, v, w> 0) $ so $ u le w ^ 3-3u $ or $ 4u le w ^ 3 $ but stagnant (I'm very bad at $ uvw $)

functional analysis – $ conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1, lambda_i ge 0, x_i in A, i = 1, cdots, n } $

Leave $ X $ be a normed vector space and $ A $ be a subset of $ X $. $ conv (A) $
It is called the intersection of all convex subsets of $ X $ that contain
$ A $

a) Show that conv (A) is a convex set

b) show that

$$ conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1,
lambda_i ge 0, x_i in A, i = 1, cdots, n } $$

c) yes $ A $ it is compact then it is $ conv (A) $ compact?

d) Show that if $ A subseteq mathbb {R} ^ n $ it is compact then $ conv (A) $ is
compact

a) Take two elements $ a $ Y $ b $ at the intersection of all convex subsets of $ X $ that contain $ A $. Now take $ lambda a + (1- lambda) b $. It is contained in each of the subsets of $ X $ that contain $ A $, therefore it is contained in the intersection. Q.E.D.

second)

$ leftarrow $:

Suppose by induction that $ sum_ {i = 1} ^ n lambda_i x_i $ It belongs to conv (A). We must prove that $ sum_ {i = 1} ^ {k + 1} lambda_i x_i $, with $ sum_ {i = 1} ^ {k + 1} = 1 $ It also belongs.

$$ sum_ {i = 1} ^ {k + 1} lambda_i x_i = sum_ {i = 1} ^ {k} lambda_i x_i + lambda_ {k + 1} x_ {k + 1} = sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1} $$

Now choose $ delta $ such that $ delta sum_ {i = 1} ^ {k} lambda_i = 1 $, so

$$ frac { delta} { delta} left ( sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1 } right = = left ( frac {1} { delta} sum_ {i = 1} ^ {k} ( delta lambda_i) x_i + frac {1} { delta} ( delta-1) x_ {k + 1} right) = \ frac {1} { delta} x + left (1- frac {1} { delta} right) x_ {k + 1} $$

which is a collection of elements of $ A $ that adds up to $ 1 $

$ rightarrow $:

I can only say that $ x in conv (A) $, so $ x = 1x + 0 cdot all $ Therefore, it is a combination that adds to $ 1 $ of elements of $ A $?

by do), give me a clue: show that conv (A union B) is the image through a continuous function of the compact $ {( alpha, beta; alpha, beta ge 0, alpha + beta = 1) } times A times B $. I do not understand how to show this.

re) Some clue?