Which one of the following inequalities is always true for positive real numbers x, y ?
[amp_mcq option1=”xy > x + y” option2=”(x + y) < (x + y)²" option3="x + y < x² + y²" option4="1 + x + y < (1 + x + y)²" correct="option4"]
This question was previously asked in
UPSC CAPF – 2016
– A) xy > x + y: If x=1, y=1, 1 > 2 is false. Not always true.
– B) (x + y) < (x + y)²: Let z = x + y. Since x, y > 0, z > 0. The inequality is z < z². This is equivalent to z² - z > 0, or z(z-1) > 0. Since z > 0, this is true only if z-1 > 0, i.e., z > 1. If x=0.1, y=0.1, x+y=0.2, which is not greater than 1. 0.2 < (0.2)² = 0.04 is false. Not always true. - C) x + y < x² + y²: If x=1, y=1, 1+1 < 1²+1² is 2 < 2, which is false. If x=0.5, y=0.5, 0.5+0.5 < 0.5²+0.5² is 1 < 0.25+0.25=0.5, which is false. Not always true. - D) 1 + x + y < (1 + x + y)²: Let w = 1 + x + y. Since x and y are positive real numbers (x>0, y>0), 1 + x + y must be greater than 1 (w > 1). The inequality becomes w < w². This is equivalent to w² - w > 0, or w(w-1) > 0. Since w > 1, both w and (w-1) are positive. Therefore, their product w(w-1) is always positive. Thus, w < w² is always true when w > 1. As 1 + x + y > 1 for positive x, y, the inequality 1 + x + y < (1 + x + y)² is always true.