For any x ∈ R, if the inequality AX2 - | x + 1 | + 2A ≥ 0 holds, then the value range of real number a is______ ．

For any x ∈ R, if the inequality AX2 - | x + 1 | + 2A ≥ 0 holds, then the value range of real number a is______ ．

The inequality AX2 - |x + 1 |x + 1 | + 2 | + 2A ≥ 0 is established, and let f (x) (f (x) = AX2 - |x + 1 |x + 1 | (f (x) = AX2 - |x + 1 |x + 1 | + 1 | (f (x) (f (x) = AX2 - |x + 1 |x + 1 | + 1 | + 2A + 2A (a \\124; f (x) (f (x) = AX2 + 2 + 2 + 2 + X + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 (x) ≥ 3+ 14}．

a. The proof of B, C ∈ (0, positive infinity): [(a + b) / a] [(B + C) / b] [(c + a) / C] ≥ 8

It is proved that: because a, B, C > 0, by the basic inequality, a + B ≥ 2 √ AB, B + C ≥ 2 √ BC, C + a ≥ 2 √ Ca, if and only if a = b = C, the above equal sign holds. So [(a + B) / a] [(B + C) / b] [(c + a) / C] ≥ (2 √ AB / a) (2 √ BC / b) (2 √ Ca / C) = 8, if and only if a = b = C

The quadratic function is known to pass through points a (0,2), B (- 1,0), C (5 / 4,9 / 8)
(1) Find the analytic expression of quadratic function
(2) Judge whether the point m (1,1 / 2) is on the straight line AC
(3) Make a straight line L through the point m and intersect the image of quadratic function at e and f (different from a, B and C). Please explore the shape of △ bef and explain the reason
The key point is the third question. 1.2 questions are all done well. The third question is, I want to know how point e comes from (it seems that all points except ABC are consistent. If so, I have to prove it),
Note: there is no picture in this problem, and the original problem has only one light coordinate system, even the image of quadratic function should be drawn by oneself

I'm a middle school teacher in Changde, Hunan Province. I've done this topic before, but I feel it's wrong~
The coordinates of point a should be (0, - 2), otherwise the calculation is very complicated
Another problem 3 should be: if the triangle bef is a right triangle, find the coordinates of point E:
(1) According to the known conditions: a = 2 & nbsp; b = 0 & nbsp; C = - 2
(2) The equation of AC is y = (& nbsp; 5 / 2) & nbsp; X-2 & nbsp; so m is on the line AC
（3）
Suppose the coordinates of point e are (P, q) & nbsp; and the coordinates of point F are (R, t), then: & nbsp;
If the triangle bef is a right triangle, then be ^ 2 + BF ^ 2 = EF ^ 2
BE^2＝(-1-P)^2+(0-Q)^2
BF^2=(-1-R)^2+(0-T)^2
EF^2=(P-R)^2+(Q-T)^2
So: (- 1-p) ^ 2 + (0-Q) ^ 2 + (- 1-r) ^ 2 + (0-t) ^ 2 = (P-R) ^ 2 + (Q-T) ^ 2... (I)
And because e and F are on parabola, so
P^2-2=Q.(i i)
R^2-2=T.(i ii)
And because
T-q / & nbsp; (R-P) = 5 / 2 & nbsp; (E, F, m are collinear). (I & nbsp; iii)
There are four equations from I to II, four unknowns, and the equations have solutions,
The coordinates of point e are obtained as (- 1 / 2, - 3 / 2)

If A-B + C = 0.8, how much is 25 (b-a-c) equal?

∵ b-a-c =-(a-b+c)
And ∵ A-B + C = 0.8
The original formula = 25 × (- 0.8)
=-20