If a b c is known, verify a 2b b 2c c 2a ab 2 bc 2 ca 2

a^2b+b^2c+c^2a-ab^2-bc^2-ca^2a^2(b-c)+a(c^2-b^2)+bc(b-c)a^2(b-c)-(ab+ac)(b-c)+bc(b-c)(b-c)(a^2-ac-ab+bc)

b-c)[a(a-c)-b(a-c)]

b-c)(a-b)(a-c)

Because a>b>c, b-c>0, a-b>0, a-c>0, so (b-c)(a-b)(a-c)>0, i.e., a 2b+b 2c+c 2a-ab 2-bc 2-ca 2>0, so a 2b+b 2c+c 2a>ab 2+bc 2+ca 2

According to the question, the difference can be obtained:

a|＜|b|Celebration Reform|c|

Combined with a b basal grip c can be obtained.

0 b c then b = -2, c = -3

A has two possible values, a = 1

L-AL=1, L-BL=2, L-CL=3, envy and old or.

a>b>c, a=1, b=a2, c=a3, or a=a1, b=a2, c=a3.

By mean inequality: a2 (a-b).

a-b)2 root number [a 2 (a-b)*(a-b)].

b-c)2 root number [b2 (b-c)*(b-c)].

2b.The above two formulas are added together.

a^2/(a-b)+b^2/(b-c)+(a-b)+(b-c)2a+2b.

So. a^2/(a-b)+b^2/(b-c)a+2b+c.

a³+b³≠6（a-1）²+2

6（a-1）²+2≥2

a³+b³≠2

This simply doesn't hold true.

A + b = 2, then a + b 2 and a + b 2, then a + b ≠ 2 is not a mutually inverse proposition, because a + b < = 2 does not necessarily lead to a + b must be equal to 2

If a+b 2 always has a +b ≠2, and cannot get a +b < = 2 must have a +b =2, this proof is wrong in principle.

The method of counterproof and the negative proposition are not equivalent, and the method of counterproof is the correct solution.

Because a>b>c,|a|=3、|b|=2、|c|=1 When c = 1, b must be 2 and a must be 3

When c = -1, if b = -2 then c = 3, if b = 2 c or = 3 then a, b, c may be 3, 2, 1

The values of a-b+c may be 2,0,4

It should be: a 2 + b 2 + c 2 = 1:

Since a+b+c=1, then (a+b+c) 2=1 so a 2+b 2+c 2+2ab+2bc+2ac=1

And because a 2 + b 2 + c 2 = 1, then ab + bc + ac = 0, so ab + c (a + b) = 0, and a + b = 1-c

ab=c^2-c.

We get ab=c 2-c, and a+b=1-c, and use Veda's theorem to get a, b is the root of two unequal real numbers of the equation x 2+(c-1)x+c 2-c=0. Therefore, its discriminant formula is greater than zero, that is, (c-1) 2-4(c 2-c)>0, and the solution is -1 30, then a>b>c>0, then ab+ac+bc>0 contradicts it, so c<0

In summary, -1 3

Take advantage of the Cauchy inequality.

a Cereb (b+c) + b (a+c)+c (a+b)] b+c)+(a+c)+(a+b)]

a (b+c) (b+c))+b (a+c) (a+c))+c (a+b) (a+b))] take the equal: a (b+c) =b (a+c) =c rent (a+b), i.e. a=b=c)).

on the inequality is.

a²/(b+c)+b²/(a+c)+c²/(a+b)]×2(a+b+c)]>a+b+c)²

a²/(b+c)+b²/(a+c)+c²/(a+b)>=a+b+c)/2

Agree with the upstairs. So the answer should be.

Because b=4a+c=8... So.

2b=a+c

So the sail beard. 2sinb=sina+sin

c because a=2c,,,a+b+c=pi

Simplify: 2sin3c=sin2c+sinc and resimplify: de.

cosc = or.

Because a>b>c,|Get.

cosc=3/4

Find it out again. sinc, and then find sina and sinb due to time problems:

sinc=root slick car number 7 4

sina = 3 root number 7 8

sinb = 5 root number 7 16

and then by the sine public trust destroys the formula.

a=12/5

c=8/5

Tsatta blind a = 2 c

b=4a+b=8 gives a=4, i.e., a= b=2 c, so 5 c=180 degrees, c=36 degrees, a=72 degrees.

4/sin72=c/sin36.Then use a calculator to look up the trigonometric value of the special angle defeat and the empty can be calculated.

From a>b>c, and a, b, c into the difference is the same as the travel knowledge: a-b=b-c, that is, 2b=a+c

and a=2cb=4 into (1) to get c=2b 3=8 3,a=2c=16 3

So the year is a = 16 3, c = 8 3

(a-b) 2=a 2-2ab+b 2>=0, in the same way (b-c) 2>=0

a-c)^2>=0

Add it up to put ab

The BCAC can be moved to the right for about a minute.