Example 4. Let x, y ∈ R, prove that | x + y | = | x | + | y | holds if and only if XY ≥ 0

Example 4. Let x, y ∈ R, prove that | x + y | = | x | + | y | holds if and only if XY ≥ 0

Sufficient: if the xy = 0, then, if the xy = 0, if the xy = 0, then, if the xy = 0, then, if the xy = 0, then, if the xy = 0, then, if the xy = 0, then, if the x = 0, then, if the x = 0, then, if the x = 0, if the xy = 0, if the xy = 0, if the xy = 0, if the xy = 0, then, if x = 0, x = 0, ② x = 0, y = 0, y = 0, x = 0, x = x + X + y = x + y = x + y = x + y = x 124124124\\\124\\\124whenxy ≥ 0, | x + y | = | x | + | y |. Necessity: obtained from | x + y | = | x | + | y |, and X, y ∈ R (x + y) 2 = (| x | + | y |) 2, that is, X2 + 2XY + y2 = x2 + 2 | XY | + Y2, then | XY | = XY, so XY ≥ 0, so the necessity is tenable. In conclusion, the original proposition is tenable, so the conclusion is tenable

Let x, yr, prove: | x + y | = | x | + | y |, if and only if XY > = 0

Solution 1: x + Y / = / X / + / Y / (x + y) ^ 2 = x ^ 2 + 2 * / XY / + y ^ 2 2XY = 2 * / XY / xy = / XY / XY > = 0. From the above, we can see that | x + y | = | x | + | y | holds if and only if XY > = 0. Solution 2: prove: 1. Sufficiency: if both sides of | x + Y / = / X / + / Y / square simultaneously, we can get: (x + y) ^ 2 = x ^ 2 + y ^ 2 + 2 * / XY / can be transformed into