Knowing that A, B, C are all unequal positive real numbers, verify B C A A A B C B C C 3

b+c-a) a+(a+b-c) b+(a+b-c) c>3 should be (b+c-a) a+(a+c-b) b+(a+b-c) c>3 is proof: (b+c) a+(a+c) b+(a+b) c>6 proof: b+c) a+(a+b) b+(a+b) cb a+c a+a b+c b+a c+b c( b a+a b)+(c b+b c)+(a c+c a) because a, b,c are all unequal positive real numbers.

b/a+a/b>2

c/b+b/c>2

a/c+c/a>2

So ( b a+a b)+(c b+b c)+(a c+c a)>6 thus (b+c-a) a+(a+c-b) b+(a+b-c) c>3

The title may be wrong, is it verified: (b+c-a) a+(a+c-b) b+(a+b-c) c>3

If this is the case, it is easy to prove it by using the basic inequality after passing the score. Equivalent to proof.

Verification: (b+c) a+(a+b) b+(a+b) c>6

Are you sure you made a mistake in transcribing the question? How do I think it's weird.

Then (b+c-a) a+(c+a-b) b+(a+b-c) c=(a+b+c)(1 a+1 b+1 c)-6 (1+1+1) 2-6=3, and the equation holds only when a 2=b 2=c 2. a, b, c are not all equal positive numbers, then the equation does not hold. ==》 （b+c-a）/a+（c+a-b）/b+（a+b-c）/c>3。

Solution 2: If x,y are both 0, then (x y)+(y x)=(x 2+y 2) (xy) (2xy) (xy) 2 (b+c-a) a+(c+a-b) b+(a+b-c) c (b a)+(c a)-1+(c b)+(a b)-1+(a c)+(b c)-1 [b a)+(a b)]+c a)+(a c)]+c b)+(b c)]-3 2+2+2-3 a, b,c is an incongruent positive number (refer to the top) 3 i.e.: (b+c-a) a+(c+a-b) b+(a+b-c) c 3

Remember to adopt it.

b+c-a) a+(a+c-b) b+(a+b-c) c is calculated to be equal to (-a-b-c)(-a-b-c)>0 Because a, b, and c are all positive real numbers, (-a-b-c)(-a-b-c)>3

To prove (b+c-a) a+(a+c-b) b+(a+b-c) c 3 is to prove (b+c) a-1+(a+c) b-1+(a+b) c-1 3 b a+a b+a c+c a+b c+c b 6 because a, b, c 0, and are not complete, so b a+a b 2 a c+c a 2 b c+c b 2 when the above equation is added, The equal sign cannot be taken because it is not fully equivalent. Therefore, the proposition b a+a b+a c+c a+b c+c b 6 is proven.

First of all, it should be (b+c-a) a+(c+a-b) b+(a+b-c) c>3

The proof is as follows. b+c-a)/a+(a+b-c)/b+(a+b-c)/c>3

It should be (b+c-a) a+(a+c-b) b+(a+b-c) c>3 is proof: (b+c) a+(a+c) b+(a+b) c>6 proof: b+c) a+(a+b) b+(a+b) cb a+c a+a b+c b+a c+b c( b a+a b)+(c b+b c)+(a c+c a) because a, b,c are all unequal positive real numbers.

b/a+a/b>2

c/b+b/c>2

a/c+c/a>2

So ( b a+a b)+(c b+b c)+(a c+c a)>6 thus (b+c-a) a+(a+c-b) b+(a+b-c) c>3

It should be greater than that.

Let , a is less than or equal to b less than or equal to c, then the equation on the left is greater than or equal to, (b+c-a) c+(c+a-b) c+(a+b-c) c is equal to (a+b+c) c is greater than or equal to 3

Proof: It should be (b+c-a) a+(c+a-b) b+(a+b-c) c>3

b+c-a)/a+(c+a-b)/b+(a+b-c)/c=b+c)/a-a/a+(c+a)/b-b/b +(a+b)/c-c/c

b+c) a-1+(c+a) b-1 +(a+b) c-1b a+c a+c b+a b+a c+b c-3(b a+a b) + (c a+a c) + (c b+b c) -3a, b, c, d are not exactly equal positive numbers.

b a+a b)>2 (b a*a b)=2c a+a c)>2 (c a*a c)=2c b+b c)>2 (c b*b c)=2 thus (b+c-a) a+(c+a-b) b+(a+b-c) c>2+2+2-3=6-3=3

b+c-a)/a+(c+a-b)/b+(a+b-c)/c>3

After a long time, I finally checked it, and it turned out that you wrote it wrong. Sweat.

b+c-a) a+(c+a-b) b+(a+b-c) c, which is ( b a+a b)+(c b+b c)+(a c+c a)-3

Because: b a+a b>2

c/b+b/c>2

a/c+c/a>2

So this equation is greater than 3.

a, b, and c are all unequal positive real numbers.

b/a+a/b>2

c/b+b/c>2

a/c+c/a>2

is the key, nothing else.

If there is a problem with the question, it should be given 3

Prove that (b+c-a) a+(a+c-b) b+(a+b-c) c>3 proves that b a+a b+c a+a c+c b+b c 6 and abc are all unequal positive real numbers.

then b a+a b 2, c a+a c 2, c b + b c 2 then b a + a b + c a + a c + c b + c 6

a, b, c are all unequal, ba

With ab, ca

with AC, CB

It is not equal to BA at all with BC

ab＞2，ca+a

c＞2，cb+b

C 2 is obtained by adding the three formulas, Ba+C

a+cb+a

b+ac+b

c＞6∴(ba+ca

1)+(cb+a

b?1)+(ac+b

c?1)＞3

i.e. b+c?a

a+a+c?b

b+a+b?cc＞3