It is known that in △ ABC, ad is the midline on the edge of BC 2（AB+AC）．

It is known that in △ ABC, ad is the midline on the edge of BC 2（AB+AC）．

It is proved that: BD + ad > AB, CD + ad > AC,
∴BD+AD+CD+AD＞AB+AC．
∵ ad is the center line on the edge of BC, BD = CD,
∴AD+BD＞1
2（AB+AC）．

Solving the quadratic inequality of one variable 1-x

I'll write the process directly
1-x1-x
3x²+x-1>0
x²+x/3-1/3>0
(x+1/6)²>13/36
x+1/6>±（√13）/6
x> (√ 13-1) / 6 or x > (- √ 13-1) / 6
Take the larger value as the range
Then the value range of X is
x>(√13-1)/6

Mathematics elective course 2-2: (1) please use synthesis method and analysis method to prove inequality: 2 root sign 2-root sign 7 < root sign 6-root sign 5

It means √ 8 - √ 7 < √ 6 - √ 5, that is √ 8 + √ 5 < √ 6 + √ 7. (√ 8 + √ 5) ^ 2 = 13 + 2 √ 40 < 13 + √ 42 = (√ 6 + √ 7) ^ 2. The square of inequality is positive, so the original equation holds?
Consider the function y = √ X. when x increases, the increment of this function becomes smaller and smaller, so √ 8 - √ 7 < √ 7 - √ 6 < √ 6 - √ 5. Is this an analytical method?

It is known that a, B, C are totally unequal positive real numbers. This paper proves that B + C − a a+a+c−b b+a+b−c c＞3．

∵ a, B, C are not equal,
∴b
A and a
b，c
A and a
c，c
B and B
C is not equal at all
∴b
A+a
b＞2，c
A+a
c＞2，c
B+b
c＞2
Add the three formulas to get B
A+c
A+c
B+a
B+a
C+b
c＞6
∴(b
A+c
a−1)+(c
B+a
b−1)+(a
C+b
c−1)＞3
B + C − a
a+a+c−b
b+a+b−c
c＞3

It is proved that the three inner angles of the triangle ABC are a, B and C respectively （1）cosA+cos(B+C)=0 （2）sin(B+C)/2=cosA/2

(1) Because the three inner angles of the triangle ABC are a, B, C, so a + B + C = 180 ° cos (B + C) = cos (π - a) = - cosa, cosa + cos (B + C) = cosa cosa cosa = 0 (2) because the three internal angles of triangle ABC are a, B, C, so a + B + C = 180 °, B + C = 180 ° - A, then sin (B + C) / 2 = sin (π - a)

Let a < B < C, prove BC ^ 2 + Ca ^ 2 + AB ^ 2 < B ^ 2C + C ^ 2A + A ^ 2B 11

The inequality moves from the left to the right
(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)
(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)
=ab(a-b)+c(b^2-a^2)+c^2(a-b)
=(a-b)(ab-c(a+b)+c^2)
=(a-b)[a(b-c)-c(b-c)]
=-(a-b)(b-c)(c-a)>0
So it was established

If a, B, C ∈ R and ab + BC + Ca = 1, then the following inequality holds A.a²+b²+c²≥2 B.(a+b+c)²≥3 C.1/a+1/b+1/c≥2√3 D.a+b+c＜√3

If a, B, C ∈ R and ab + BC + Ca = 1, then the following inequality holds
.a²+b²+c²≥ab+bc+ac=1
(a+b+c)²=a²+b²+c²+2(ab+bc+ac)》3（ac+bc+ab)=3
So choose
B

In order to know three non-zero rational numbers a, B.C., find the value of a △| a | + B | B | + C | C | + | ab | AB + | BC | BC + | Ca + | ABC | I'm a new user,

a/|a|+b/|b|+c/|c|+|ab|/ab+|bc|/bc+|ca|/ca+|abc|/abc
It is divided into the following situations:
First, all three numbers are positive
Original formula = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7
Second, all three numbers are negative
The original formula = - 1-1-1 + 1 + 1 + 1-1 = - 1
Third, two are positive and one is negative
Original formula = 1 + 1-1 + 1-1-1-1 = - 1
Fourth, two are negative and one is positive
The original formula = - 1-1 + 1 + 1-1-1 + 1 = - 1

1. If a square plus b square plus C square minus AB minus BC minus CA is equal to 0, it is proved that a = b = C 2. Given a square * b square + a square + b square + 1 = 4AB, find the values of a and B

1、
a²+b²+c²-ab-bc-ac=0
Double 2 on both sides
2a²+2b²+2c²-2ab-2bc-2ac=0
(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)=0
(a-b)²+(b-c)²+(c-a)²=0
If the square is greater than or equal to 0, the sum is equal to 0. If there is one greater than 0, at least one is less than 0, which is not true. So all three are equal to 0
So A-B = 0, B-C = 0, C-A = 0
a=b,b=c,c=a
So a = b = C
2、
a²b²+a²+b²+1=2ab+2ab
(a²b²-2ab+1)+(a²-2ab+b²)=0
(ab-1)²+(a-b)\x06=0
If the square is greater than or equal to 0, the sum is equal to 0
If one is greater than 0, the other is less than 0
So both are equal to zero
So AB-1 = 0, A-B = 0
A=b
Substituting AB = 1
a²=1
a=±1
So a = 1, B = 1 or A-1, B = - 1

Find the smallest positive real number k so that the inequality AB + BC + Ca + K (1 / A + 1 / B + 1 / C) is greater than or equal to 9 for all positive real numbers a, B, C

From the problem design and the basic inequality x + y + Z ≥ 3 (XYZ) ^ (1 / 3), we can get
ab+[k/(2a)]+[k/(2b)]≥3(k²/4)^(1/3)
bc+[k/(2b)]+[k/(2c)]≥3(k²/4)^(1/3)
ca+[k/(2c)]+[k/(2a)]≥3(k²/4)^(1/3)
By adding the above three formulas, we can get
ab+bc+ca+[(1/a)+(1/b)+(1/c)]≥9(k²/4)^(1/3)
According to the question design, there must be 9 (k? 2 / 4) ^ (1 / 3) ≥ 9
 there must be K ﹣ 4
∴k≥2
The minimum value of K is 2