Find 50 examples of math inequalities in the second year of high school! 30

Multiply both sides of equation 1 (a+1)+1 (b+1)=1 by (a+1) (b+1), simplify the left and right equations to get ab=1, from the basic inequality x+y>=2 root number xy, we can get a+2b>=2 root number a2b, according to ab=1, a+2b>=2 root number 2, therefore, the minimum value is 2 root number 2

(1) Equivalent to mx2-2x+(1-m) 0 for any real number x constant, divided into m=0 and m≠0 two cases to discuss, and then use greater than 0 constant conditions to be satisfied: the opening is upward, and the discriminant formula is less than 0 to solve the value range of m

2) The equivalence of (x2-1)m-(2x-1) 0 is constant on [-2,2], and the use of a function can be discussed separately in the case of either an increasing function or a subtracting function

Answer: Solution: (1) The original inequality is equivalent to mx2-2x+(1-m) 0 for any real number x.

When m=0, -2x+1 0 x$ frac$ is inconsistent.

\left\\\endight.$,m has no solution, so m does not exist

2) Let f(m)=(x2-1)m-(2x-1).

To make f(m) 0 constant on [-2,2], if and only if.

left\\\endight.$⇔left\^-2x-1＜0}\\2x+3＜0}\endight.$

\frac}＜x＜\frac}$

x can be in the range of } x frac}$}

Comments: This question examines the constant establishment problem of the primary function and the quadratic function The constant establishment problem of the quadratic function is divided into two categories, one is greater than 0, the constant formation must meet the opening upward, and the discriminant formula is less than 0, and the other is less than 0, the constant formation must satisfy the opening downward, and the discriminant formula is less than 0

x-x-2 0 gives x<-1 or x>2

2x +(2k+5)x+5k=(x+k)(2x+5) 0 solves -k-5 2, and since there is only an integer solution of -2, x<-1 is solved in front

So -2<-k -1

So in summary, 1 k<2

x -x-2=(x-2)(x+1) 0 --x<-1 , or x>2 is the set of integer solutions, so x<-1 is taken

2x²+（2k+5)x+5k=(2x+5)(x+k)<0 --5/2k<2

The value range of the real number k is k<2

(1)a²/b +b≥2a

b²/c +c≥2b

c²/a +a≥2c

The above 3 formulas are added together.

a b+b+b c+c+c a+a 2a+2b+2c(a b+b c+c a)+(a+b+c) 2(a+b+c) so a b+b c+c a a+b+c(2) proves: b 2 ab

a+c≥2√a

c+b≥2√bc

Multiply the three formulas to obtain:

a+b)(b+c)(a+c)>=8ab ac bc=8abc[(a+b) 2][(a+c) 2][(a+c) 2]>=abc and take the logarithm at the same time to obtain:

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

1.Proof: a b+b c+c a a+b+c

Proof: (a b+b c+c a) ·(b+c+a).

√(a²/b·b) +b²/c·c + c²/a·a))²

a+b+c)²

a²/b+b²/c+c²/a ≥ a+b+c

Method 2: a b+b c+c a)+(a+b+c).

a²/b+b) +b²/c+c)+ c²/a+a)

2√[(a²/b)·b]+2√[(b²/c)·c]+2√[(c²/a)·a]

2a+2b+2c

2(a+b+c)

So a b+b c+c a a+b+c, the original inequality is proven.

2.Due to the existence of the bottom change formula, there is no difference between LN and LG.

a+b)/2+㏑(b+c)/2+㏑(c+a)/2 = ln[(a+b)(b+c)(c+a)/8]

a+㏑b+㏑c = ln(abc)

So the inequality that needs to be proved is (a+b)(b+c)(c+a) 8abc

a+b≥2√ab,b+c≥2√bc，c+a≥2√ca

So (a+b)(b+c)(c+a) 2 ab·2 bc·2 ca = 8abc

Thus the proposition is proven.

Definition of one positive, two definite and three equal:

Refers to the special requirements that are specified and emphasized when proving or solving a problem with inequality a+b 2 ab.

One positive: both a and b must be positive;

Two certain: 1When a+b is a fixed value, the maximum value of a*b can be known;

2.When a*b is a fixed value, you can know the minimum value of a+b;

Three equals: The equal sign is true if and only if a and b are equal; That is, at a b, a+b 2 ab is 1 x>0 because x>0, so the maximum value can be found by the mean inequality. x*(1 x)=1 is a fixed value, so when x=1 x, there is a minimum value.

The equation x=1 x is multiplied by x at the same time, x 2=1, and x > 0, and x=1 is solved

One positive, two definite and three equal refers to the special requirements specified and emphasized when proving or solving a problem with the inequality a+b 2 ab.

One positive: both a and b must be positive;

Two certain: 1When a+b is a fixed value, the maximum value of a*b can be known;

2.When a*b is a fixed value, you can know the minimum value of a+b;

Three equals: The equal sign is true if and only if a and b are equal; That is, in a b, a+b 2 ab in your problem, x is equivalent to a, 1 x is equivalent to b, x>0, 1 x>0 is equivalent to satisfying a positive, so the theorem says that when a b, a+b is the smallest, and the minimum value is 2 ab, in your problem, when x=1 x, f(x) is the smallest, and the minimum value is 2

This makes it easier when it comes to solving inequalities with its practical images.

The first thing to do is to understand the basic inequality.

a+b)/2 >= √(ab)

A positive ab is a positive number to ensure that both sides are meaningful (negative numbers cannot be greater than positive numbers, and cannot be negative numbers under the root number).

The second certainty is that once A+B or AB has a fixed value, AB can have a maximum value, and A+B can have a minimum value, and the three equals are when A=B, and the equal sign is true.

To prove it is simple, a-b) 2>=0

a^2 -2ab +b^2 >=0

A 2 + 2AB +B 2 >= 4AB-> A+B) 2>=4AB (Root on both sides)->A+B)>=2 (AB)->A+B) 2 >= (AB) Therefore, in the above question, A=X B=1 X, AB=1, according to the two determinations, A+B>=2

When to take the equal sign? Three phases: a=b i.e. x=1 x. Because x>0, x=1

Let's see if you understand something?

Good luck with your studies!

Answer: One positive, two definite and three equal.

For example, a+b>=2, the root number ab is positive, that is, a b is positive at the same time; The second definite is that a+b or ab is the fixed value; Three equals is"="The condition for the establishment is that the equal sign is established when a=b.

For example, x 0, f(x)=x+1 x in 1 x is positive, x=1 x is the equal sign is the condition for the equal sign, so it can be deduced that when x=1 or -1 is true, the equal sign is true, and because x 0 is x, x can only be equal to 1

So f(x)=x+1 x>=2, root number, 1 x, take the minimum value when x can only be equal to 1, i.e., f(x)=2 times 1 = 2

One positive: both a and b must be positive;

2. Definite: When a*b is a fixed value, you can know the minimum value of a+b;

Three equals: The equal sign is true if and only if a and b are equal; i.e., at a b, a+b2ab proves: (a- b) 0 a+b-2 ab 0 a+b 2 ab

When ab is a fixed value, a+b has a minimum value, and if and only if (a- b) = 0, i.e., a=b, there is a minimum value of 2 ab. The reason why it has to be positive is because when it is negative, a and b don't make sense in high school math.

This question should first be discussed on a case-by-case basis! What does it mean that the symbol does not change? It's always positive or negative! Then the steps to solve the problem will come out naturally!

I do it according to the idea of combining numbers and shapes!

First of all, when f(x) > 0, just make f(-2) >0 and f(2) >0! (because it is a one-time function, it is monotonic), and the solution is <

1/x+5x/y

3x+4y)/5x+5x/y

3/5+(4y/5x+5x/y)

3/5+2√(4y/5x▪5x/y)

Only if 4y 5x = 5x y, 4y = 25x , {2y = 5x, {3x + 4y = 5, the solution gives x=5 13, y=25 26 takes the equal sign, so the minimum value is 23 5.

The substitution method is relatively straightforward and easy to think of, you can turn the objective function into a univariate function, and then use the derivative to find the maximum value, but if the objective function is complex and not easy to simplify, it is necessary to solve it with the help of the method of inequality

You're right, it's not true that the inequality is used twice and the conditions are different.

This problem needs to use the basic inequality after the transformation, and the direction of the transformation is to reduce the formula with a reciprocal relation, and then use the basic inequality to reduce the x and y terms to get the minimum value.

Hope it helps!

I think it's 4 and 3 5. I do it by substitution of "1".

The length of the bus bar of the cylindrical lantern is L, which is known from the title.

l = , so the area of the material used is s=s side + s bottom = -3 (, so when r =, s obtains a maximum value of about square meters.

Four congruent rectangular skeletons, the total consumption of meters of wire, then a rectangular skeleton consumes meters of wire, if the length of the rectangular skeleton is x, then the width is.

s=π×（x/2)²+x×（

When x=, s has a maximum value, and the radius is .