If the real numbers A, B, C satisfy A B C, A B C 0, AB BC CA 0, ABC 0

1-abc>0 may be 2 negative and one positive, all positive.

2-a>b>c combine 1 to illustrate a>0, so b<0, c<03-a+b+c>0 illustrate a>b+c

4-ab+bc+ca<0 indicates that a is a positive real number, b and c are negative real numbers, and ab+ca<0, ab+ca

Because abc is greater than 0, there are only two cases where it is greater than two positive and one negative out of three.

and ab+bc+ac is less than 0, so it can only be case 2.

Because a is greater than b is greater than c, so a is greater than 0 and b is greater than c, but b and c are both less than 0 and a+b+c is greater than 0, so |a|Greater than |b|+|c|Therefore, |a|Greater than |b|。Select (c).

ab+bc+b+c^2+16=0

b(a+c+1)+c^2+16=0

a+c+1=b+8 (according to the merger conditions, the source skin can be obtained) is substituted to obtain b 2 + 8 b + 16 + c 2 = 0

b+4)^2+c^2=0

b=-4,c=0

a=3b a=-4 Crack vertical 3

Answer] Scatterer: Cave Calendar A

The main knowledge points examined in this question are the properties of unequal rushing potato style

a>b, then a-c>b-c

(1) Let a be maximum, and there must be a 0, b+c=2-a, bc=4 a, then b,c are the two real roots of the equation x 2-(2-a)x+4 a=0.

then =(a-2) 2-4*4 a 0

Removing the denominator yields a 3-4a 2+4a-16 0,(a-4)(a 2+4) 0

So a 4 is again a = 4 and b = c = -1

That is, the minimum value of the largest in a, b, and c is 4

2) Because abc = 4 0, a+b+c = 2 0

So a, b, and c may all be positive, or one positive and two negative.

When a, b, and c are all positive, the minimum value of the largest of a, b, and c is known from (1) is 4, which contradicts a+b+a=2.

When a, b, and c are positive and negative, let a 0, b 0, and c 0

then |a|+|b|+|c|=a-b-c=a-（b+c）=a-（2-a）=2a-2

From (1) to know a 4

So 2a-2 6

So |a|+|b|+|c|The minimum value is 6

Add a>0 by a+b>c, a+c>b, and then note that there is at least one positive number in b and c.

If b and c are both positive, then a+b>a,b+c>b,c+a>c, multiply to get the conclusion.

If b=1, (a+b)(b+c)(c+a>0>=abc is obtained by a+b>c>00,b+c>0,c+a>0

If c0>=abc

Solution: by a+b+c=

0 shows that two of the three numbers a, b, and c must be negative, and abc 0

1 is a positive value, x=1-1-1=-1

a+b+c=0

3abc,y=-3

x+2y+3xy

Because a>b>0 , a-b>0 is known to be a 2-2ab+b 2=2ab , i.e. (a-b) 2=2ab , and a 2+2ab+b 2=6ab , so (a+b) 2=6ab , so [(a-b) (a+b)] 2=(2ab) (6ab=1 3 , so (a-b) (a+b) = 3 3 .

ab<0 indicates a and b variant signs.

AC>0 indicates that A and C have the same sign.

When a>0, then b<0 and c>0

then abc <0

Original = 1-1 + 1-1 = 0

When a>0, then b<0 and c>0

At this time, ac>0 (negative negative is positive), and because b>0, so abc is also 0 because of |abc|>0, so abc |abc|=1 original = -1+1-1+1=0

In summary, the original formula = 0

ps: The answer is not wrong, there is only one result, 1354753447 this landlord wrote it wrong, I said it very clearly above, you can think about it yourself.

AB is different, AC is the same sign, AC is the same is positive.

a/|a|+b/|b|+c/|c|=1，abc/|abc|=-1，a/|a|+b/|b|+c/|c|+abc/|abc|=0ac when both are negative.

a/|a|+b/|b|+c/|c|=-1，abc/|abc|=-1，a/|a|+b/|b|+c/|c|+abc/|abc|=-2

Solution: If a>0, then b<0 and c>0. So the original formula = 1-1 + 1-1 = 0

If a<0, then b>0 and c<0. So the original formula = 1-1 + 1-1 = 0