2 votes 2 votes Let $x, y$ be two positive numbers such that $x + y =1.$ Then, the minimum value of $\left(x+ \frac{1}{x} \right)^2 + \left(y + \frac{1}{y}\right)^2$ is $12$ $20$ $12.5$ $13.3$ Quantitative Aptitude cat2001 quantitative-aptitude algebra + – go_editor asked Mar 31, 2016 • edited Apr 10, 2022 by Lakshman Bhaiya go_editor 13.9k points 1.8k views answer comment Share See all 0 reply Please log in or register to add a comment.
0 votes 0 votes for getting minimum value x will be 1/2 , y =1/2 So, that x+y=1 then value of (x+(1/x))^2 +(y+(1/y))^2 =12.5 Answer will be 3) srestha answered Apr 1, 2016 srestha 5.2k points comment Share See all 2 Comments See all 2 2 Comments reply Arjun 8.6k points commented Apr 3, 2016 reply Share Why x = y = 0.5? 1 votes 1 votes Soumik Dutta 10 points commented May 1, 2023 reply Share Same question: Why x=y=½ for getting min value? And what if Max value is asked? 0 votes 0 votes Please log in or register to add a comment.