A (1-A) less than or equal to one fourth? If so, please prove

Yes. A (1-A) ≤ 1 / 4
Certificate:
a(1-a)
=a-a^2
=-a^2+a
=-a^2+a-1/4+1/4
=(1/4)-(a-1/2)^2
The square term is constant and nonnegative, (A-1 / 2) ^ 2 ≥ 0
(1/4)-(a-1/2)^2≤(1/4)-0=1/4
So a (1-A) ≤ 1 / 4

An = 2n-5, BN = an / 2 ^ n, let the sum of the first n terms of BN be TN, it is proved that 1 / 4 is greater than or equal to TN and less than 1

There is a problem in the title BN = an / 2 ^ NbN = (2n-5) / 2 ^ NTN = B1 + B2 + +bn=(2-5)/2+(2*2-5)/2^2+… +（2n-5）/2^n…… .12Tn=(2-5)+(2*2-5)/2+(2*3-5)/2^2+…… (2n-5) / 2 ^ (n-1). Formula 22-1 gives TN = - 3 + 2 / 2 + 2 / 2 ^ 2 + +2/2^(n-1)-(2n-5)/2^n=-3-(2n...

N is greater than or equal to 3, TN = the nth power of 3 - (n + 3) (1 / 2). Compare the sizes of TN and 5N / (2n + 1) and prove that

TN is larger. Since n = 4, TN is larger than 2.5 (it can be proved that TN is an increasing sequence). However, 5N / (2n + 1) is always less than 2.5

Who can prove why one plus one equals two?

First of all, the conjecture is not to prove 1 + 1 = 2, but to prove that an even number greater than 2 can be decomposed into the sum of two prime numbers. Second, what Chen proves is 1 + 2, not 1 + 1. What Chen proves is that an even number greater than 2 can be decomposed into the sum of the product of one prime number and the other two prime numbers. 1 + 1 = 2, which is called axiom

A (1-A) less than or equal to one fourth? If so, please prove