Inequalities comparing two linear moduli

For k>0, both sides of k|u| <= |v| are nonnegative. Squaring gives the equivalent inequality k2u2 <= v2. Move the squared expressions to one side; factor the difference of squares before expanding when useful. Determine the product sign and include equality endpoints. |x-1| <= |x+1| gives (x-1)2 <= (x+1)2, hence x 0. It compares distances to 1 and -1. Check: do not require either inner expression to be positive. If leading squared terms cancel, solve the resulting linear inequality rather than forcing a quadratic formula.