Can you explain the inequality of average value? I am a senior three, and I can't remember the knowledge of high school clearly.

Average inequality

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1, harmonic mean: HN = n/(1/a1/a2+...+1/an) 2, geometric mean: GN = (A 1A2 ... an). Square average: qn = √ (a12+a22+...+an 2)/n The formula that these four averages satisfy Hn≤Gn≤An≤Qn is the average inequality.

catalogue

Brief introduction of mean inequality

Deformation of mean inequality

Proof of mean inequality

Application of mean inequality

Other inequalities

Important inequality-1. cauchy inequality

Important inequality -2. Unequal ranking

Important inequality -3. chebyshev inequality

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Brief introduction of mean inequality

Concept:

1, harmonic average: HN = n/(1/a1/a2+...+1/pieces)

2. Geometric mean: gn = (A 1A2 ... ampere) (1/n)

3. Arithmetic average: An=(a 1+a2+...+an)/n

4. Square average: qn = √ [(a12+a22+...+an2)/n]

These four averages satisfy Hn≤Gn≤An≤Qn.

A 1, a2, …, an∈R+, if and only if a 1=a2= … =an, take "=".

General form of mean inequality: let function d (r) = [(a1r+a2r+... anr)/n] (1/r) (when r is not equal to 0);

(a 1a2...an) (1/n) (when r=0) (that is, d (0) = (a 1a2 ... an) (1/n))

Then there is: when R noticed that Hn≤Gn≤An≤Qn is only a special case of the above inequality, that is, D (-1) ≤ D (0) ≤ D (1) ≤ D (2).

A simple conclusion can be drawn from the above simplification: 2/(1/a+1/b) ≤√ AB ≤ (a+b)/2 ≤√ [(A2+B2)/2] is commonly used in middle schools.

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Deformation of mean inequality

(1) for real numbers a and b, there exists a 2+b 2 ≥ 2ab (if and only if a=b, take "=", a 2+b 2 > 0 >: -2ab

(2) For nonnegative real numbers A and B, a+b≥2√(a*b)≥0, that is, (a+b)/2≥√(a*b)≥0.

(3) For negative real numbers A and B, A+B.

(4) For real numbers A and B, a(a-b)≥b(a-b).

(5) For non-negative numbers A and B, a 2+b 2 ≥ 2ab ≥ 0.

(6) For nonnegative numbers A and B, A 2+B 2 ≥ 1/2 * (A+B) 2 ≥ 2AB.

(7) For nonnegative numbers A, B and C, there exists a 2+B 2+C 2 ≥1/3 * (A+B+C) 2.

(8) For nonnegative numbers A, B and C, there exists a 2+B 2+C 2 ≥ AB+BC+AC.

(9) For nonnegative numbers A and B, there exists a 2+AB+B 2 ≥ 3/4 * (A+B) 2.

(10) There is (a+b+c)/3 > for real numbers A, B and C; =(abc)^( 1/3)

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Proof of mean inequality

There are many methods, such as mathematical induction (first or later induction), Lagrange multiplier method, Qinsheng inequality method, rank inequality method, Cauchy inequality method and so on.

Proved by mathematical induction, an auxiliary conclusion is needed.

Lemma: let A≥0 and B≥0, then (a+b) n ≥ an n+na (n-1) b.

Note: The correctness of lemma is obvious. Conditions A≥0 and B≥0 can be weakened to A≥0 and A+B ≥ 0. Interested students can think about how to prove it (by mathematical induction).

The original title is equivalent to: ((a1+a2+…+an)/n ≥ a1a2 … an.

When n = 2, it is easy to prove;

Suppose that the proposition holds when n = k, that is,

((a 1+a2+…+AK)/k)^k≥a 1a2…ak。 Then when n = k+ 1, let A (k+ 1) be the largest of a 1, a2, …, A (k+ 1), then

k a(k+ 1)≥a 1+a2+…+ak .

Let s = a 1+A2+…+AK,

{[a 1+a2+…+a(k+ 1)]/(k+ 1)}^(k+ 1)

= { s/k+[k a(k+ 1)-s]/[k(k+ 1)]}^(k+ 1]

≥ (s/k) (k+1)+(k+1) (s/k) k [ka (k+1)-s]/k (k+1) by lemma.

=(s/k)^k* a(k+ 1)

≥a 1a2…a(k+ 1). Use inductive hypothesis.

Here is a simple way to understand.

Qinsheng inequality method

Qin Sheng inequality: convex function f(x), x 1, x2, ... xn is any n points of function f(x) in the interval (a, b),

Then it is: f [(x1+x2+...+xn)/n] ≥1/n * [f (x1)+f (x2)+...+f (xn)]

Let f(x)=lnx, and f(x) is a convex increasing function.

Therefore, ln [(x1+x2+...+xn)/n] ≥1/n * [ln (x1)+ln (x2)+...+ln (xn)] = ln [(x/kloc-0)]

That is, (x1+x2+...+xn)/n ≥ (x1* x2 * ... * xn) (1/n).

Prove by the theorem of projection in a circle (the radius is not less than half a chord)

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Application of mean inequality

Example 1 Prove inequality: 2 √ x ≥ 3-1/x (x >); 0)

Proof: 2 √ x+1/x = √ x+1/x ≥ 3 * [(√ x) * (1/x)] (1/3) = 3.

Therefore, 2 √ x ≥ 3-1/x.

Example 2 The area of a rectangle is p, so find the minimum perimeter.

Solution: let the length and width be a and b respectively, then a * b = p.

Because a+b≥2√(ab), 2 (A+b) ≥ 4 √ (AB) = 4 √ P.

The minimum circumference is 4 √ p.

Example 3 The circumference of a rectangle is p, and the maximum area is found.

Solution: Let the length and width be a and b respectively, then 2 (a+b) = p.

Because a+b=p/2≥2√(ab), AB ≤ P 2/ 16.

The largest area is p 2/ 16.

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Other inequalities

Zhan Sen inequality

Absolute inequality

Weight and inequality

Holder inequality

minkowski inequality

bernoulli inequality

cauchy inequality

chebyshev inequality

Waisenbik inequality

Sequential inequality

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Important inequality-1. cauchy inequality

Cauchy inequality is generally proved as follows:

The formal writing of (1) Cauchy inequality is: if the number of two columns is AI and BI, there is (∑ AI 2) * (∑ BI 2) ≥ (∑ AI * BI) 2.

Let f (x) = ∑ (ai+x * bi) 2 = (∑ bi 2) * x 2+2 * (∑ ai * bi) * x+(∑ ai 2).

Then we know that there is always f(x) ≥ 0.

Using the condition that the quadratic function has no real root or only one real root, there are δ = 4 * (∑ AI * Bi) 2-4 * (∑ AI 2) * (∑ Bi 2) ≤ 0.

So move the item to the conclusion.

(2) Prove by vector.

m=(a 1，a2......an) n=(b 1，b2......bn)

Mn = a1b1+a2b2+...+anbn = (a1+a2+...+an) 1/2 times (b 1+B2+ ...

CosX is less than or equal to 1, so: a1b1+a2b2+...+AnBN is less than or equal to a1+a2)12 times (b65438+.

This proves inequality.

There are many kinds of Cauchy inequality, and only two commonly used proofs are taken here.

Cauchy inequality is a common theoretical basis for finding the maximum value of some functions and proving some inequalities, which should be paid great attention to in teaching.

Clever disassembly constant:

Example: Let A, B and C be positive numbers and not equal.

Proof: (2/a+c)+(2/b+c)+(2/c+a) > (9/a+b+c)

Analysis: ∫a, B and C are all positive numbers.

∴ Only need to prove that the conclusion is correct: 2 (a+b+c) [(1/a+b)+(1/b+c)+(1/c+a)] > 9.

And 2(a+b+c)=(a+b)+(a+c)+(c+b).

The other 9 = (1+1+1) (1+1)

Prove: θ 2 (a+b+c) [(1/a+b)+(1/b+c)+(1/c+a)] = [(a+b)+(b+)

A, b and c are not equal, so the equal sign cannot be established.

∴ The original inequality holds.

There are many examples like this, which are not listed in the entry. You can find specific documents about the proof and application of Cauchy inequality in resources.

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Important inequality -2. Unequal ranking

Rank inequality is the basic inequality required by the outline of high school mathematics competition.

There are two groups of numbers A 1, A 2, ... A N, B 1, ... B N satisfies A 1 ≤ A 2 ≤...A N and B 1 ≤ B 2 ≤...≤ B N has A/kloc-0. 1+...+a nb1≤ a1b t+a2bt+...+a nbt ≤ a1b1+a2b2+...+nbn where t 1.

The above rank inequality can also be simply described as: the sum of inverse order and ≤ disordered order and ≤ congruence order.

The gradual adjustment method can be used in the proof.

For example, it is proved that if a 1 b 1+a 2 b 2 is adjusted to a 1 b 2+a 2 b 1, the value will become smaller, and only the difference (A1-A2) * (b/kloc) is needed.

By analogy, according to the gradual adjustment method, the ranking inequality is proved.

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Important inequality -3. chebyshev inequality

There are two Chebyshev inequalities.

(1) Suppose that the sequence A 1, A2, A3 ... Ann and B 1, B3 B2...BN satisfy A 1 ≤ A2 ≤ A3 ≤ ... Ann and B 1 ≤ B2 ≤ B3 ≤ ...

Then, ∑ Abby ≥( 1/n)(∑ai)(∑bi)

(2) Let the existence sequences A 1, A2, A3 ... An and B 1, B3 B2...BN satisfy A 1 ≤ A2 ≤ A3 ≤...an and B 1 ≥ B2 ≥ B3...≥ BN.

Then, ∑ Abby ≤( 1/n)(∑ai)(∑bi)

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Important inequality -4. Qinsheng inequality

Let f(x) be a convex function, then f [(x1+x2+...+xn)/n] ≥ [f (x1)+f (x2)+...+f (xn)]/n is called piano inequality (power average).

The weighted form is:

F [(a1x1+a2x2+...+anxn)] ≥ a1f (x1)+a2f (x2)+...+anf (xn), where

Ai>= 0 (I = 1, 2, ... n) and a 1+A2+...+An = 1.

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Important inequality -5. Average inequality

A 2+b 2 ≥ 2ab (the sum of squares of a and b is not less than twice their product).

When a and b are greater than 0 respectively, the upper measurement can be changed to a+b ≥2√ab.

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Important inequality -6. Complete mean inequality

√[(a^2+ b^2)/2]≥(a+b)/2≥√ab≥2/( 1/a+ 1/b)

(Quadratic power average ≥ arithmetic average ≥ geometric average ≥ harmonic average)

Proof: (The proof process is quoted from him)

Let a and b be two positive numbers,

m2=√[(a^2+b^2)/2],a=(a+b)/2,g=√(ab),h=2/( 1/a+ 1/b)

Represent the quadratic power average, arithmetic average, geometric average and harmonic average of A and B respectively. Prove: m2 ≥ a ≥ g ≥ h.

It is proved that in trapezoidal ABCD, AB‖CD, AB=b, CD = a. EiFi(i= 1, 2, 3, 4) is a line segment parallel to the bottom of trapezoidal ABCD, which is cut by two waists of the trapezoid.

If E 1F 1 is divided into two parts and the products are equal, then

e 1f 1=√[(a^2+b^2)/2]。

If E2F2 is divided into trapezoidal midlines, then

E2F2=(a+b)/2 .

If E3F3 is divided into two similar graphs, then

E3F3=√(ab)。

If E4F4 passes through the line segment of the intersection of two diagonal lines of the trapezoid, then

E4F4=2/( 1/a+ 1/b)。

It is intuitively proved from the diagram that e1f1≥ e2f2 ≥ e3f3 ≥ e4f4, and when a=b, the equal sign is taken.

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Important inequality -7. Power mean inequality

Power means inequality: ai & gt0( 1≤i≤n), and α >; β, then (∑ AI α/n)1/α≥ (∑ AI β/n)1/β holds.