High school a basic inequality problem! 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

High school a basic inequality problem! 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

x=y=1/3
Then 1 / x + 1 / Y ≠ 2 / 3
So obviously not
According to your algorithm, it should be 2 √ 1 / (XY) = 6
But a + b > = 2 √ (AB)
The right side must be a fixed value
And here √ 1 / XY is not a fixed value
So we should use (2x + y) (1 / x + 1 / y) here
The result is 3 + 2 √ 2

Basic inequality in Senior High School
If ab ≠ 0, a, B ∈ R are known, then ()
A.b/a+a/b≥2 B.b/a+a/b≥-2 C.b/a+a/b≤-2 D.|b/a+a/b|≥2

If there is a negative number in A. B, the inequality will not hold

The problem of basic inequality in Senior High School
0 ＜ a ＜ B, the following inequality is correct
A a ＜ B ＜ radical ab ＜ (a + b) / 2
BA ＜ radical ab ＜ (a + b) / 2 ＜ B
Ca ＜ radical ab ＜ B ＜ (a + b) / 2
D radical ab ＜ a ＜ (a + b) / 2 ＜ B

∵b＞a＞0;
∴a+b＞2√ab
∴(a+b)/2＞√ab
Obviously, B ＞ (a + b) / 2 ＞ ab ＞ a;
Select B;