Given that a + B + C = 0 and ABC is not equal to 0, calculate the value of a {1 / B + 1 / C} + B {1 / C + 1 / a} + C {1 / A + 1 / b} + 3,

Given that a + B + C = 0 and ABC is not equal to 0, calculate the value of a {1 / B + 1 / C} + B {1 / C + 1 / a} + C {1 / A + 1 / b} + 3,

a{1/b+1/c}+b{1/c+1/a}+c{1/a+1/b}+3
=a/b+a/c+b/c+b/a+c/a+c/b+3
=[a(b+c)+b(a+c)+c(a+b)]/abc+3
=(2ab+2ac+2bc)/abc+3.①
∵a+b+c=0;
∴(a+b+c)(a+b+c)=a*a+b*b+c*c+2ab+2ac+2bc=0;
∴2ab+2ac+2bc=0
∴①=0/abc+3
=3
The original formula is 3
Violent solution
Take a = - 1; b = - 1; C = 2 to calculate

It is known that | A-1 / 2 | + | B + 1 / 3 | + | C + 2 / 5 | = 0, (1) compare the size of ABC? (2) calculate the value of | a | + | (- 6) | + | C |
What are the values of a respectively?
（1）｜a｜＝－a　（2）｜a｜＝｜－a｜ （3）a／｜a｜＝1　　　（4）｜a｜/a=-1

A is greater than B is greater than C
|The value of a | + | - 6 | + | C | is equal to 6.9

a. B. C is the length of three sides of triangle ABC, calculate the value of √ (a = b = C) ^ 2 + √ (a-b-c) ^ 2 + √ (b-a-c) ^ 2 + √ (c-b-a) ^ 2

According to the trilateral relationship, there are
a-b-c＜0;b-a-c＜0;c-b-a＜0;
be
√（a+b+c)^2 +√（a-b-c)^2 +√（b-a-c)^2 +√（c-b-a)^2
=（a+b+c)+（a-b-c)+（b-a-c)+（c-b-a)
=0.