Concept: The arithmetic mean of n positive real numbers is greater than or equal to its geometric mean.
Arithmetic average, the average value obtained by dividing the sum of a set of numbers by the set of numbers. Also called average.
Geometric average, the n-th root of a number is the product of n factors, usually the square root of n.
Let a 1, a2, a3, ..., an be n positive real numbers, then (a 1+a2+a3+...+an)/n≥n times √(a 1*a2*a3*...*an), if and.
● Deformation of average 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 ≥ AB.
(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.
● Proof of mean inequality ——————————————————————
There are many methods, including mathematical induction (first or later induction), Lagrange multiplier method, Qin Sheng inequality method, rank inequality method, Cauchy inequality method and so on.
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, then f' (x) =1/x >; 0,f"(x)=- 1/x^2<； 0, so f(x) is a convex increasing function.
Therefore, ln [(x1+x2+...+xn)/n] ≥1/n * [ln (x1)+ln (x2)+...+ln (xn)] = lnn times √ (x650).
That is, x 1+x2+...+xn≥n times √(x 1*x2*...*xn).
● Application of mean inequality ————————————————————————————————————————————————————————————————————.
Example 1 Prove inequality: 2 √ x ≥ 3-1/x (x >); 0)
Proof: 2 √ x+1/x = √ x+1/x ≥ 3 * 3 times √ (√ x) * (1/x) = 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.
● Summary of mean inequality ————————————————
1, mixed average: HN = n/(1/a1+1/A2+...+1/each)
2. Geometric mean: gn = (a1a2 ... an) (1/n) = n times √(a 1*a2*a3*...* an).
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 "=".
In a triangle, the sum of two sides must be greater than the third side, which is the triangle inequality.
Trigonometric inequality is very simple, but it is the most basic conclusion in plane geometric inequality, including generalized Ptolemy theorem, euler theorem inequality and Euler inequality, which will be used to derive inequality relations in the end.
The trigonometric inequality also has the following inference: two intersecting line segments AB and CD must have AC+BD less than AB+CD.
|a|-|b| less than or equal to |a+b| less than or equal to |a|+|b| (theorem), also known as trigonometric inequality.
The inequality with an unknown number and the highest degree of the unknown number is called quadratic inequality, and its general form is AX 2+BX+C > 0 or AX 2+BX+C.
The solution of the unary quadratic inequality 1) When v ("v" means yes, the same below) = b 2-4ac > =0, the quadratic trinomial, ax 2+bx+c has two real roots, then ax 2+bx+c can always be decomposed into a (x-x1). In this way, solving a quadratic inequality can be reduced to solving two linear inequalities. The solution set of unary quadratic inequality is the union of the solution sets of these two unary linear inequalities.
For example.
2x^2-7x+6<； 0
Use cross multiplication.
2x -3
1x -2
Get (2x-3) (x-2) < 0
Then, it is discussed in two situations:
I. 2x-3
Get x; 2。 wrong
Second, 2x-3 & gt；; 0，x-2 & lt； 0
Get x> 1.5 and x
The solution set of the final inequality is: 1.5
In addition, the collocation method can also be used to solve quadratic inequalities:
2x^2-7x+6
=2(x^2-3.5x)+6
=2(x^2-3.5x+3.0625-3.0625)+6
=2(x^2-3.5x+3.0625)-6. 125+6
=2(x- 1.75)^2-0. 125<； 0
2(x- 1.75)^2<； 0. 125
(x- 1.75)^2<； 0.0625
Both sides are squares, and you must
x- 1.75 & lt； 0.25 and x- 1.75 >: -0.25
X<2 and x> 1.5
The solution set of inequality is 1.5.
The unary quadratic inequality can also be solved by the unary quadratic function image. By looking at the image, we can know the two intersections between the quadratic function image and the X axis, and then deduce the answer according to the "< 0" or "> 0" required by the topic.
To find the solution set of a univariate quadratic inequality is actually to move all the terms of this univariate quadratic inequality to the left of the inequality and discuss factorization and classification to solve the set. Solving the unary quadratic inequality can transform the unary quadratic equation inequality into the form of quadratic function, find out the intersection of the function and the X axis, connect the unary quadratic inequality, quadratic function and unary quadratic equation, and solve the problem by mirror method, which simplifies the problem.
Basic properties of common inequalities: a>b, b>c =>a & gtc；;
a & gtb = & gta+c & gt； b+ c；
A>b, c>0 = & gtac & gt BC;
A>b, c<0 = & gtac & lt BC
; a & gtb & gt0，c & gtd & gt0 = & gtac & gtBD；
a & gtb，ab & gt0 = & gt 1/a & lt； 1/b
; A>b>0 =>a power of>A > n power of B;
Basic inequality: (radical ab)≤(a+b)/2
Then it can be changed to Party A-2AB+Party B ≥ 0.
Party A+Party B ≥ 2ab
There are two!
One is ||| A |-| B || ≤| A-B |≤| A |+| B | B |
The other is |||| A |-| B ||≤| A+B |≤ A |+| B |
Prove that you can use vectors. A and B are regarded as vectors, and the difference between two sides of a triangle is smaller than the third side.
The sum of two sides is greater than the third side.
Cauchy inequality:
Let a 1, a2, …an, b 1, b2…bn all be real numbers, then (a1b1+a2b2+…+anbn) 2 ≤ (a12+a22+).
The essence of inequality:
Add (or subtract) the same algebraic expression on both sides of 1. inequality, and the direction of inequality remains unchanged.
2. Both sides of the inequality are multiplied by (or divided by) the same positive number, and the direction of the inequality remains unchanged.
3. When both sides of the inequality are multiplied by (or divided by) the same negative number, the direction of the inequality will change.
4. When both sides of the inequality are multiplied by 0, the inequality becomes an equal sign.