The two questions of high school inequality are relatively simple, and I don t know how to simplify

aˇ2+√2a+1)(aˇ2-√2a+1)=[(aˇ2+1)+√2a][(aˇ2+1)-√2a]=(a^2+1)^2-2a^2；

aˇ2+1)+a][(aˇ2+1) -a]=(a^2+1)^2-a^2;Since (a 2+1) 2 is always positive and subtracts the positive number, the result of subtracting the smaller one is larger, so the second one is larger.

If the exam doesn't require a process, you can choose to substitute the number, which is generally constant, for example, if x=0, then the first one is 1, and the second one is 1 2, and the result is easy to get.

a 2+ 2a+1)(a 2- 2a+1)=[(a 2+1)+ 2a]*[a 2+1)- 2a], the first question is similar to this;

x+1)(x 2+x 2+1)=(x+1)[(x 2+x+1)-x 2], respectively, the second question can be done in this way.

The main thing is to make up formulas.

It is recommended that you take a look at the formulas for (a+b) 2 and (a+b) 3.

Basic inequality problem types and solutions: Solve absolute value problems (simplification, evaluation, equations, inequalities, functions), and transform problems with absolute values into problems without absolute values.

1) Classification discussion method: remove the absolute value according to the number in the absolute value sign or the first positive, zero and negative fraction of the formula.

2) Zero-point segmented discussion method: It is suitable for the case of multiple random cultivation of absolute values containing one letter.

3) Two-sided flat method: suitable for equations or inequalities that are non-negative on both sides.

4) Geometric meaning method: It is suitable for situations with obvious geometric significance.

Two big tips. 1". If the sum of two formulas is a constant, the minimum value of the sum of the reciprocals of these two formulas is usually multiplied by 1, and then 1 is expressed by the previous constant, and the two formulas can be calculated.

If the sum of the reciprocals of the two formulas is known to be constant, find the minimum value of the sum of the two formulas, and the method is the same as above.

The basic inequality question types and solutions are as follows:1. After making a difference, the difference is obtained by decomposing the factors and formulas to determine the sign of the difference.

2. Quotient (algebraic formula often used for the power of fractional exponents) analytical method leveling method; The numerator (or denominator) is physicochemical; Use the monotonicity of the function to find the middle or deflation method) image method jujube argument.

3. Among them, the comparative method (making a difference, making a quotient) is the most basic method.

Precautions: 1. Symbols:

1. Add or subtract the same number or formula on both sides of the inequality, and the direction of the inequality sign remains unchanged.

2. Multiply or divide the two sides of the inequality by the same positive number, and the direction of the inequality sign remains unchanged.

3. Multiply or divide the two sides of the inequality by the same negative number, and the direction of the inequality sign changes.

2. Solution set: 1. If it is larger than both values, it is larger than the larger one (the same is larger).

2. If it is smaller than both values, it will be smaller than the smaller one (the same as the smaller).

3. Carry auspicious than the big one, smaller than the small one, there is no solution (the big and the small can't be taken).

4. Larger than small, smaller than large, there is a solution in the middle (small and large take the middle).

5. An inequality group composed of three or more inequalities can be inferred.

3. Number axis method:

The solution set of each inequality is represented on the number line, and the points on the number line divide the number line into segments, and if a certain segment of the number line represents the solution set on the number line is the same as the number of inequalities, then this section is the solution set of the inequality group. A few will cost a few.

When determining a quadratic inequality, a>0, δ=b 2-4ac>0, the solution set of inequalities is available"It is greater than both sides and smaller than the middle"Find.

1. About the inequality: ("equivalent to hail" is denoted as """) xy 2 1 4 (x 2 + y 2).

Multiply by 42xy x 2+y 2

The additional attack was awarded 2xy

4xy x 2+y 2+2xy=(x+y) 2 divided by 8xy 2 (x+y) 2 beats 8

xy/2≤1/4（x^2+y^2） xy/2

It is known that a b c is a positive number.

Verification: (1) A 2B 2+B 2C 2+A 2C 2 is greater than or equal to ABC(A+B+C).

2) (b+c-a) a+(c+a-b) b+(a+b-c) c≥3

3) LG[(A+B) 2]+LG[(B+C) 2]+LG[(C+A) 2] is greater than or equal to LGA+LGB+LGC

Answer: (1).

Make use of the formula: x 2 + y 2 + z 2 > = xy + yz + xz

a^2b^2 + b^2c^2 + a^2c^2

ab)^2 + bc)^2 + ac)^2

ab*bc + ab*ac + bc*ac

abc(a + b + c)

Use the formula: x + y > = 2 (xy).

b+c-a) a + c+a-b) b + a+b-c)

b/a + c/a + c/b + a/b + a/c + b/c - 3

b/a + a/b) +c/a + a/c) +c/b + b/c) -3

Modular Kai = 2 + 2 + 2 - 3

Use the formula: x + y > = 2 (xy).

a+b)/2 >=ab)

So: lg[(a+b)2] >lg[ ab)].

lga + lgb ) 2

Similarly: LG[(B+C) 2] >LGB + LGC) 2

lg[(c+a)/2] >lga + lgc ) 2

Add the three formulas to get:

lg[(a+b)/2] +lg[(b+c)/2] +lg[(c+a)/2] >lga + lgb + lgc

The inequality of the same number can only be added by shouting Qimin, and can not be done by subtracting the number method, such as the addition of the two formulas to obtain a>7 2, and the subtraction of the two formulas is equivalent to the addition of a + b>3 and -a + b<-4, of course, it is not possible, pay attention to the inequality when the two sides of the inequality are multiplied by a negative number, the unequal sign should be changed. The solution to this problem is to multiply the first inequality by the positive number 2 (unchanged sign) and then add the second inequality.

Two inequalities can only be added to each side if the same is greater than or the same is less than.

It's like the first sock loose 2>1; 2>1 Do these two subtraction equations still hold up for the calendar?

This should be the content of linear programming, the method is to establish a direct training trap angle coordinate system, the two axes are a, b axis, one straight line is a+b-3 0, the other is a-b-4 0Then use the shadow to represent the same range, and then set Z=3a+b, get B=3A+Z, and translate this line in the coordinate system, you can clearly see the value range of Z.

As for why it will only expand, you can draw a diagram and compare the area.