Let the real numbers a, B and C form an equal proportion sequence, and the non-zero real numbers x and y are the median of the equal difference between a and B, B and C respectively, and prove that ax + CY = 2

Prove: because a, B, C are equal ratio sequence, so & nbsp; & nbsp; B2 = AC ① and X, y are equal difference median of a and B, B and C respectively, so & nbsp; & nbsp; & nbsp; & nbsp; 2x = a + B, 2Y = b + C ② to prove & nbsp; & nbsp; & nbsp; ax + CY = 2 as long as prove & nbsp; & nbsp; ay + CX = 2XY as long as prove & nbsp; & nbsp

It is known that non-zero real numbers a, B and C form an arithmetic sequence, and the tolerance D ≠ 0. It is proved that 1 / (√ B + √ C), 1 / (√ C + √ a), 1 / (√ a + √ b) also form an arithmetic sequence

Suppose D > 0, then c > b > A, 1 / (√ B + √ C) = (√ C - √ b) / (√ B + √ C) * (√ C - √ b) = (√ C - √ b) / (C-B) = (√ C - √ B) / D, similarly: 1 / (√ C + √ a) = (√ C - √ a) / 2D; 1 / (√ a + √ b) = (√ B - √ a) / d. let them subtract each other to get (√ C - √ a) / 2D - (√ C - √ b) / D = (2 √ B - √ a - √ C) / 2D, (√ B - √ a) / D - (√ C - √ a) / 2D = (2 √ B - √ a - √ C) / 2D, i.e. (√ C - √ a) / 2D - (√ C - √ b) / D = (√ B - √ a) / D - (√ C - √ a) / 2D, i.e. 1 / (√ a + √ b) - 1 / (√ C + √ a) = 1 / (√ C + √ a) - 1 / (√ B + √ C), 1 / (√ C + √ a), 1 / (√ a + √ b) also form an arithmetic sequence

If two different non-zero real numbers a, B, C form an arithmetic sequence, and a, C, B form an equal ratio sequence, what is a / b equal to

a. B and C are non-zero real numbers, and they are not equal to each other, a ≠ B
a. B, C into arithmetic sequence
2b=a+c
c=2b-a
a. C, B are equal ratio sequence
c^2=ab
(2b-a)^2=ab
The result is a ^ 2-5ab + 4B ^ 2 = 0
(a-b)(a-4b)=0
A = B (rounding off) or a = 4B
a/b=4

Given the real number a B C, satisfy a + B + C = 0 and ABC = 0, prove that at least one of a B C is not less than 2
Sorry, the title is wrong, should be: known real number a B C, satisfy a + B + C = 0 and ABC = 2, verify: A, B, C at least one is not less than 2.

Let AB be negative and C be positive
A. The absolute value of B is less than C
Suppose ABC is less than 2, then the product of ABC must be less than 2
So if it doesn't hold, ABC can't all be less than 2
That is, at least one of a, B and C is not less than 2

Let the real numbers a, B and C form an equal proportion sequence, and the non-zero real numbers x and y are the median of the equal difference between a and B, B and C respectively, and prove that ax + CY = 2