It is proved that if x and y are positive real numbers and X + y ＞ 2, at least one of 1 + XY ＜ 2 or 1 + YX ＜ 2 holds

It is proved that if x and y are positive real numbers and X + y ＞ 2, at least one of 1 + XY ＜ 2 or 1 + YX ＜ 2 holds

It is proved that neither 1 + XY < 2 nor 1 + YX < 2 holds, that is, 1 + XY ≥ 2 and 1 + YX ≥ 2 (2 points) ∵ x, y are all positive numbers, ∵ 1 + X ≥ 2Y, 1 + y ≥ 2x (5) 1 + X + 1 + y ≥ 2x + 2Y (8) x + y ≤ 2 This is in contradiction with the known x + y ＞ 2 The hypothesis is not true, that is, 1

Take a counter example to show that the proposition "for any real number x, the value of x square + 6x + 8 is greater than 0" is a false proposition!

If x is a real number, then (x + 3) 2 > 0, the counterexample to prove that it is a false proposition is ()
A. X = 0b. X = - 3C. X = 3D. There is no counterexample

If (x + 3) 2 ＞ 0, then x + 3 ≠ 0, the solution is x ≠ - 3. Therefore, we can prove that "x is a real number, then (x + 3) 2 ＞ 0" is a false proposition, and the counterexample is x = - 3