Comparison method is the most basic method to prove inequality, including difference comparison method and quotient comparison method. The basic idea is to change the formula that is difficult to compare into the difference between it and 0 or the comparison size between it and the quotient of 1. When the two ends of the proved inequality are partial (or fractional), it is often used for difference comparison, and when the two ends of the proved inequality are in product form (or power exponential form is often used for quotient comparison)
Example 1: a+b≥0 is known, and it is proved as: a3+b3≥a2b+ab2.
Analysis: From the observation of the topic, we can compare the differences, then extract the common factor, and explain the positive or negative after the differences with a+b≥0, so as to prove the inequality. The step is 10, make a difference of 20, and sort out the positive and negative differences of 30 judgments.
∵(a3+b3)？ (a2b+ab2)
=a2(a-b)-b2(a-b)
=(a-b)(a2-b2)
Proof: =(a-b)2(a+b)
∫(a-b)2≥0a+b≥0
∴(a-b)2(a+b)≥0
That is, a3+b3≥a2b+ab2.
Example 2 let a, b∈R+ and a≠b, and prove: aabb>abba.
Analysis: According to the verified inequality, A and B have rotational symmetry, so they can be used as quotient comparison method on the premise that A > B > 0, and the quotients can be compared with "1" to achieve the purpose of proof. The steps are as follows: 10 quotient 20 quotient ranking 30 is judged as the size of 1.
It is proved that the law of A > B > 0 can be solved from the symmetry of a and b.
aabbabba=aa-b？ bb-a=(ab)a-b
∵a？ b？ 0,∴ab？ 1，a-b？ 0
∴(ab)a-b？ (ab)0= 1 means AABBABA > 1, ABBA > 0 ∴ AABB > ABBA.
Exercise 1 known a, b∈R+, n∈N, proved (a+b) (an+bn) ≤ 2 (an+1+bn+1).
Basic inequality method
It is a common method to prove inequality by using basic inequality and its variants. Commonly used basic inequalities and variants are:
(1) if a, b∈R, a2+b2≥2ab (if and only if a=b, take the equal sign).
(2) if a and b∈R+, then a+b≥ 2ab (if and only if a=b, take the equal sign).
(3) If the symbols of A and B are the same, then ba+ab≥2 (if and only if a=b, take the equal sign).
Example 3 if a, b∈R, | A | ≤ 1 and | B | ≤ 1, then a1-B2+b1-a2≤1.
Analysis: It can be directly applied through observation: xy≤x2+y22.
Prove: ∫ a1-B2B1-A2 ≤ A2+(1-B2) 2+B2-(1-A2) 2 =1.
∴ b1-a2+a1-B2 ≤1,the equal sign is true if and only if a 1+b2= 1.
Exercise 2: If A? b？ 0, proving that a+ 1(a-b)b≥3.
synthetic method
The synthesis method is to calculate the inequality to be proved from the known or proved inequalities according to the nature of the inequality.
Example 4, suppose a? 0，b？ 0, a+b= 1, which proves that: (a+1a) 2+(b+1b) 2 ≥ 252.
Proof: ∫a? 0，b？ 0，a+b= 1
∴ab≤ 14 or 1ab≥4
left = 4+(A2+B2)= 1 A2+ 1 B2 = 4+[(A+B)2-2AB]+(A+B)2-2 ABA 2 B 2。
= 4+( 1-2ab)+ 1-2 ABA 2 B2≥4+( 1- 12)+8 = 252
Exercise 3: It is known that a, b and c are positive numbers, n is a positive integer, and f (n)= 1gan+bn+cn3.
Verification: 2f(n)≤f(2n)
analyse
Starting from the theory, we find a sufficient condition for the proposition to hold, and only when this condition is a provable or proven inequality can we deduce that the original inequality holds. This method is called analytical method.
Example 5: Known A? 0，b？ 0，2c？ A+b, verification: c-C2-ab < a < c+C2-ab.
Analysis: It is observed that the verification formula is a chain inequality, so it is not convenient to use comparison method. It is also observed that the verification formula is equivalent to | a-c | < C2-ab, and the basic inequality method is not applicable, but the analysis method is more applicable.
Prove that c-C2-ab < a < c+C2-ab.
Just prove that -C2-AB < A-C < C2-AB.
Proof: namely | a-c | < C2-ab.
Direct evidence (a-c) 2 < C2-ab
Evidence a2-2ac
∫a > 0, ∴ means a-2c.
This inequality still exists.
Exercise 4: a∈R, a≠ 1 is known. Prove: 3 (1+A2+A4) > (1+A+A2) 2.
Scaling method
Scaling method is an important method to prove inequality by properly enlarging or narrowing one side of inequality and using the transitivity of inequality. The commonly used skillful techniques are: (1) discarding some positive terms (or negative terms), (2) enlarging (or narrowing) some terms in the sum or product, and (3) enlarging (or narrowing) the score.
Example 6: It is known that A, B, C and D are all positive numbers.
Verification:1< ba+b+c+CB+c+d+DC+d+a+ad+a+b < 2.
Analysis: Observing the characteristics of the formula, if the quotient of the four fractions is the same denominator, the problem can be solved. If the quotient is the same denominator, you can also use the scaling method, but the common denominator is too troublesome, so use the scaling method.
Prove that ∫ Ba+B+C+CB+CB+D+DC+D+A+AD+A+B > Ba+B+C+D+CA+B+C+D+DA+B+C+D+AA+B+C+D = A+B+C.
And from ab < a+MB+m (0 < a < b, m > 0), we can get: ba+b+c < b+da+b+c+DC b+c+d < c+aa+b+c+DDC+d+a < d+BC+d+a+dad.
∴ba+b+ c+c b+ c b+ c b+ d+DC+d+a+ad+a+b < b+ da+b+ c+d+c+aa+b+ c+d+d+BC+d+a+d+a+d+a+ca+b+ c+d = 2(a+b+ c+c+c)a+b+ c+d = 2
To sum up:1< ba+b+CB+CB+c+d+DC+d+a+ad+a+b < 2.
Exercise 5: known: A < 2, proved: LOGA (A+ 1) < 1.
6 alternative methods
Substitution method can play a role in solving many practical problems. Some problems are difficult to prove directly. It is very convenient to solve them through method of substitution's thoughts and methods. It is often used to prove conditional inequalities, and trigonometric substitution is common.
(1) triangle replacement:
It is a common replacement method. When solving algebraic problems, we use proper trigonometric functions for substitution, and transform algebraic problems into trigonometric problems, and make full use of the properties of trigonometric functions to solve problems.
Example 7, if x, y∈R+, x-y= 1? A = (x-1y) (y+1y).1x, verification 0 < a < 1.
Prove: ∫x, y∈R+, and x-y= 1, x=secθ, y=tanθ, (0 < θ < xy).
∴a =(secθ- 1 secθ(tanθ+ 1 tanθ 1 sec 2θ
= 1-cos 2θcosθS2 m2θ+cos 2θcosθS2θcos 2θ
= sine θ
∵ 0 < θ < x2, ∴ 0 < s2mθ < 1 so 0 < a < 1.
Review 6: Known 1≤x2+y2≤2, verified: 12 ≤x2-xy+y2≤3.
(2) Ratio substitution:
For the problem that there are several equal proportions in the known conditions, an auxiliary unknown can be set to represent this proportion, and then it can be substituted into the verification formula.
Example 8: x- 1=y+ 12=z-23 is known. Verification: x2+y2+z2≥43 14.
Proof: let x-1= y+12 = z-23 = k.
So x=k+ 1, y=zk- 1, and z=3k+2.
Substitute the above formula into x2+y2+z2 = (k+1) 2 (2k-1) 2+(3k+2) 2.
= 14(k+5 14)2+43 14≥43 14
reductio ad absurdum
If it is difficult to prove some inequalities clearly from the front, we can consider the reduction to absurdity, that is, first deny that the conclusion is not valid, and then gradually deduce contradictory or self-contradictory conclusions with definitions, theorems, axioms or known conditions according to known conditions, so as to confirm that the original conclusions are correct. Reduction to absurdity is applicable to all propositions that are at least, unique or contain negative words.
Example 9: p3+q3=2 is known, which proves that p+q≤2.
Analysis: This question has been known P and Q three times, but only once in the conclusion. It is difficult to prove it simultaneously with the cubic root method and the scaling method, so consider using the reduction to absurdity.
Prove: let p>2-q > 2, then p+q > 2-q.
∴p3>(2-q)3=8- 12q+6q2-q3
Substituting p3+q3 =2 gives 6q2- 12q+6 < 0.
That is, 6 (q- 1) 2 < 0, which leads to the contradiction of ∴p+q≤2.
Exercise 7: A+B+C > 0, AB+BC+AC > 0 and ABC > 0 are known.
Verification: a > 0, b > 0, c > 0.
complete induction
Inequalities related to natural number n are usually proved by mathematical induction. There are two steps to prove the problem by mathematical induction.
Example 10: Let n∈N and n > 1, and verify: (1+13) (1+15) ...
Analysis: observation and verification formula is related to n, and mathematical induction can be used.
Proof: (1) When n=2, left = 43 and right =52.
The inequality ∫43 > 52∴ holds.
(2) Assuming n=k(k≥2, k∈n), the inequality holds, that is, (1+13) (1+15) ...
Then when n=k+ 1, (1+13) (1+15) ...
Prove that the left side of formula ① is > 2k+32, as long as it is 2k+ 12.
2k+22k+ 1>2k+32②
For ② < 2 > 2k+2 > 2k+ 1 2k+3.
(2k+2) 2 > (2k+ 1) (2k+3)
< 2 > 4k2+8k+4 > 4k2+8k+3
< 2 > 4 > 3 (3)
∵ ③ holds ∴ ② holds, that is, when n=k+ 1, the original inequality holds.
It is proved by (1)(2) that the original inequality holds for all n≥2(n∈N).
Exercise 8: n ∈ nKnown, n > 1. Verification:1n+1+1n+2+…+12n >1324.
structured approach
According to the concrete structure of proving inequality, the purpose of proving inequality is achieved by constructing functions, sequences, composite numbers and graphs. This method is called construction method.
1 constructor method
Example 1 1: Prove inequality: x 1-2x < x2 (x ≠ 0).
Proof: Let f(x)= x 1-2x- x2 (x≠0).
∫f(-x)
=-x 1-2-x+x2x-2x2x- 1+x2
= x 1-2x-[ 1-( 1-2x)]+x2 = x 1-2x-x+x2
=f(x)
∴ f (the image ∴f(x) represents the y axis symmetry.
When x > 0, 1-2x < 0, so f (x) < 0.
When x < 0, we know that f (x) < 0 according to the symmetry of the image.
When x≠0, there is always f (x) < 0, that is, x 1-2x < x2 (x ≠ 0).
Exercise 9: A > B, 2B > A+C is known, and it is proved that B-B2-AB < A < B+B2-AB.
2 structure graphic method
Example 12: if f(x)= 1+x2, a≠b, then | f (x)-f (b) | | a-b |
Analysis: According to the structure of 1+x2, this is the distance between two points A( 1, x) and 0 (0,0) on the rectangular coordinate plane, that is,1+x2 = (1-0) 2+(x-0) 2.
Let A( 1, a) and B( 1, b) then 0A= 1+a2.