1. Compare the size ①a^2+b^2___ (a+b)^/2 ②ab_____ (a^2+b^2)/2 ③(a+b)^2____ 4ab ④[(a+b)/2]^2_____ (a^2+b^2)/2 2. Given x > 0, Y > 0, xy = 4, then the minimum value of X + y? 3. Given x > 0, Y > 0, x + y = 6, then the maximum value of XY? 4. Given that x > 0, then the minimum value of (32 / x) + 2x is? Trouble has a process

1. Compare the size ①a^2+b^2___ (a+b)^/2 ②ab_____ (a^2+b^2)/2 ③(a+b)^2____ 4ab ④[(a+b)/2]^2_____ (a^2+b^2)/2 2. Given x > 0, Y > 0, xy = 4, then the minimum value of X + y? 3. Given x > 0, Y > 0, x + y = 6, then the maximum value of XY? 4. Given that x > 0, then the minimum value of (32 / x) + 2x is? Trouble has a process

1. > = = 2 radical (XY) = 2 * radical 4 = 2 * 2 = 4 3. Xy = 2 radical [(32 / x) * 2x] = 2 * 8 = 16

Mathematical problems about inequality in grade one of senior high school
It is known that a belongs to R. if the equation x ^ 2 + X + | A-1 / 4 | + | a | = 0 has real roots, then the value range of a is obtained

If A-1 / 4 > = 0 and a > = 0
Then a > = 1 / 4
The original formula is x ^ 2 + X + A-1 / 4 + a = 0
x^2+x+2a-1/4=0
If the equation has real roots, then 1-4 * (2a-1 / 4) > = 0
1-8a+1>=0
a=1/4
So a = 1 / 4
If A-1 / 4 = 0
So 0

1.1/2>(1/2+1/4+...+1/2n)/n (n>=2)
2.1/(n+1)(1+1/3+1/5+,+1/(2n-1))>1/n(1/2+1/4+,+1/2n) (n>=2)

(1) Multiply both sides by n
n/2>1/2+1/4+...+1/2^n
All the molecules on the right side use two bands (e.g. 1 / 4N (1-1 / 2) = n / 2 > 1 / 2 + 1 / 4 +, + 1 / 2n) (the proof method is the same as the first question)
Get proof