a. B, C ∈ r a + B + C = 1 to prove ((1 / a) - 1) ((1 / b) - 1) ((1 / C) - 1) ≥ 8

a. B, C ∈ r a + B + C = 1 to prove ((1 / a) - 1) ((1 / b) - 1) ((1 / C) - 1) ≥ 8

There should be another condition that a, B and C are all greater than 0 ～ [otherwise, take a = b = 0.75, C = - 0.5, then the inequality will not hold ～]
It's just been working for a long time
(1 / a) - 1 = (B + C) / A is easy to see after general division, because a + B + C = 1 ～]
Similarly, (1 / b) - 1 = (a + C) / B, (1 / C) - 1 = (B + a) / C ～
Take all the three formulas in and transform them again. The equivalence of the original formula is proved
(a+b)(b+c)(a+c)≥8abc
At this time, according to the formula a + B ≥ 2 √ (AB), that √ is treated as the root, and the root can't be marked for a while ～]
There are a + B ≥ 2 √ (AB), B + C ≥ 2 √ (BC), C + a ≥ 2 √ (AC)
If you multiply the three formulas above, you will get the following result:
(a+b)(b+c)(a+c)≥8√（abcabc）=8abc

There are four points a (0,1), B (2,1), C (3,4), D (- 1,2) in the plane rectangular coordinate system. To prove that the four points are on the same circle (geometric method)

A straight line passing through the middle point of AB must pass through the center of the circle. Let m (1, m) be the coordinate of the center of the circle
∴MA=MC
That is, (1-0) & #178; + (m-1) & #178; = (1-3) & #178; + (M-4) & #178;, the solution is m = 3
The center of the circle is (1,3), Ma = MC = √ (1 & # 178; + 2 & # 178;) = √ 5
∴MB=√[(1-2)²+(3-1)²]=√5,MD=√[(1+1)²+(3-2)²]=√5
Ma = MB = MC = MD, that is, a, B, C and D are on the same circle

Given a point (1,0,1) B point (4,4,6) C point (2,2,3) d point (10,14,17), proving ABCD coplanar

Three points define a plane
CP vector is (1,0, radical 2)
The CM vector is (1 + X, 0, z)
So when 1 + x = root 2 of Z / square, or 1 + x = - Z / square of two lines
Not collinear at other times
When root, 1 + x = Z / root 2 shows collinearity because they have the same starting point in the same direction vector a (different starting points, which are parallel)
1 + x = - Z / 2 times the root of the square, the reverse collinear
The CP vector obtained by subtracting p
MCC subtracts to get centimeter vector
The equal sign is proportional to the carrier of the above two conditions

Four points a (0,0), B (10,0), C (12,6), D (2,6) on known coordinate plane
If the line y = mx-3m + 2 divides the quadrilateral ABCD into two parts with equal area, then the value of M is

It is easy to know that ABCD is a parallelogram, so the area is divided into two parts, then the line must pass through the midpoint (6,3) of the parallelogram
Substituting into the linear equation y = mx-3m + 2
M = 1 / 3