Some basic inequalities in Senior High School 1. For any x > 0, X / (the square of X + 3x + 1) ≤ a, then the value range of a is? 2. Let a > b > 0, then the minimum value of the square of a + 1 / AB + 1 / a (a-b) is?

Some basic inequalities in Senior High School 1. For any x > 0, X / (the square of X + 3x + 1) ≤ a, then the value range of a is? 2. Let a > b > 0, then the minimum value of the square of a + 1 / AB + 1 / a (a-b) is?

The following is x + 1 / x + 3. X + 1 / X is greater than or equal to 2, so a is greater than or equal to 1 / 5
2. Merge 1 / AB + 1 / a (a-b) = 1 / b (a-b), and then divide a ^ 2 by a ^ 2 under the root of 2 * (a ^ 2 / ab-b ^ 2), which becomes 1 / [B / a (1-B / a)], and then the mean inequality [B / a (1-B / a)] is less than or equal to 1 / 4, so the minimum answer is 4

Given that x + 2Y = 1, both X and y are greater than 0, find the minimum value of 1 / x + 1 / y
Because 1 / x + 1 / Y "2 √ (1 / XY)
If and only if 1 / x = 1 / y is obtained, then x = y
Substitute x + 2Y = 1 to get x = y = 1 / 3
So the minimum value is 2 / 3
Please point out the wrong steps and logical mistakes!
Good answer, add 100 points!

The correct solution is: pay attention to the condition that 1 is replaced by 1 / x + 1 / y = (x + 2Y) / x + (x + 2Y) / y = 3 + 2Y / x + X / Y > = 3 + √ (2Y / x) * (x / y) = 3 + √ 2 when taking the equal sign, 2Y / x = x / y, x + 2Y = 1 to get the value of X, y when taking the equal sign, take the equal sign directly above, without considering the following

Given A2 + B2 = 1, X2 + y2 = 1, prove ax + by ≤ 1

It is proved that: ∵ A2 + B2 = 1, X2 + y2 = 1, ∵ A2 + B2 + x2 + y2 = 2, ∵ A2 + x2 ≥ 2aX, B2 + y2 = 2by ∵ 2aX + 2by ≤ 2, ∵ ax + by ≤ 1