Let a, B, C be a three digit hundred, ten and one digit number respectively, and a ≤ B ≤ C, then the maximum value of | A-B | + | B-C | + | C-A |______ .

Let a, B, C be a three digit hundred, ten and one digit number respectively, and a ≤ B ≤ C, then the maximum value of | A-B | + | B-C | + | C-A |______ .

∵ a, B and C are the hundreds, tens and ones of a three digit number respectively, and a ≤ B ≤ C,
The minimum value of a is 1, and the maximum value of C is 9,
∴|a-b|+|b-c|+|c-a|=b-a+c-b+c-a=2c-2a，
The maximum possible value of | A-B | + | B-C | + | C-A | is 2 × 9-2 × 1 = 16
So the answer is 16

If the side length of the equilateral triangle ABC is 1, vector BC = a, vector CA = B, vector AB = C, then AB + BC + Ca is

∵ AB + BC + Ca = 0 vector
Square on both sides:
（AB+BC+CA）²=0
∴|AB|²+|BC|²+|CA|²+2AB●BC+2BC●CA+2CA●AB=0
∵ the side length of the equilateral triangle ABC is 1
∴2AB●BC+2BC●CA+2CA●AB=-3
∴BC●CA+CA●AB+AB●BC=-3/2
∴ab+bc+ca
=BC●CA+CA●AB+AB●BC
=-3/2

It is known that ABC satisfies that a + B part AB is equal to one-third part, B + C Part BC is equal to one fourth part, and C + a part CA is equal to one fifth part

ab/(a+b)=1/3
Take the reciprocal
(a+b)/ab=3
a/ab+b/ab=3
1/b+1/a=3
In the same way
1/b+1/b=4
1/a+1/c=5
Add up
2(1/a+1/b+1/c)=12
1/a+1/b+1/c=6
All points
(ab+bc+ca)/abc=6
Take the reciprocal
abc/(ab+bc+ca)=1/6

It is known that a, B and C are real numbers, and ab a+b＝1 3，bc b+c＝1 4，ca c+a＝1 5. Find ABC The value of AB + BC + Ca

Take the reciprocal of the known three fractions to get a + B
ab＝3，b+c
bc＝4，c+a
ca＝5，
That is 1
A+1
b＝3，1
B+1
c＝4，1
C+1
a＝5，
Add the three formulas to get 1
A+1
B+1
c＝6，
Results: ab + BC + ca
abc＝6，
That is, ABC
ab+bc+ca=1
6．

If ABC = 1, then a ab+a+1+b bc+b+1+c The value of Ca + C + 1 is () A. 1 B. 0 C. -1 D. -2

∵ ABC = 1, ᙽ a, B, C are not 0
A
ab+a+1+b
bc+b+1+c
ca+c+1
=ac
1+ac+c+b
bc+b+1+bc
1+bc+b
=abc
b+1+bc+b
bc+b+1+bc
1+bc+b
=1+b+bc
b+1+bc=1．
Therefore, a

Given that a, B, C are real numbers, and AB / A + B = 1 / 3, BC / B + C = 1 / 4, CA / C + a = 1 / 3, calculate the value of ABC / A + B + C

Because AB / (a + b) = 1 / 3, BC / (B + C) = 1 / 4, CA / (c + a) = 1 / 3,
Take the reciprocal and get 1 / A + 1 / b = 3, 1 / A + 1 / C = 4, 1 / C + 1 / a = 3
So 1 / a = 1,1 / b = 1 / C = 2
So the reciprocal of ABC / (a + B + C) is 4
Therefore, ABC / (a + B + C) = 1 / 4

It is known that a, B and C are real numbers, and ab a+b＝1 3，bc b+c＝1 4，ca c+a＝1 5. Find ABC The value of AB + BC + Ca

Take the reciprocal of the known three fractions to get a + B
ab＝3，b+c
bc＝4，c+a
ca＝5，
That is 1
A+1
b＝3，1
B+1
c＝4，1
C+1
a＝5，
Add the three formulas to get 1
A+1
B+1
c＝6，
Results: ab + BC + ca
abc＝6，
That is, ABC
ab+bc+ca=1
6．

Proof of inequality, 1/(n+1)+1/(n+2)+1/(n+3)+..+1/3n>4n/(4n+1)

0

Proof of inequality, Given that a, B, C belong to R +, a + B + C = 1, it is proved that a ^ 2 + B ^ 2 + C ^ 2 > = 1 / 3?

(1) (a + B + C) ^ 2 = a ^ 2 + B ^ 2 + C ^ 2 + 2Ab + 2Ac + 2BC = 1, because (2) a ^ 2 + B ^ 2 > = 2Ab (3) a ^ 2 + C ^ 2 > = 2Ac (4) B ^ 2 + C ^ 2 > = 2BC add up the left side of the five formulas 3A ^ 2 + 3B ^ 2 + 3C ^ 2 + 2Ab + 2Ac + 2BC is greater than or equal to the right side of the five formulas

Inequality proof Let f (x) = ax 2 + BX + C (a > 0, c) Please write more details`

The question is wrong, it should be C > 0, otherwise the second question is wrong
First of all, see how to use the conditions
From F (c) = 0 and Vieta theorem, another root is 1 / A
When a > 0 and 0