3 votes 3 votes If $a, b, c, d$ and $e$ are positive real numbers, then the minimum value of $(a+b+c+d+e)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e})$ is $25$ $5$ $125$ Cannot be determined. dhingrak asked Dec 6, 2014 edited Mar 1, 2017 by Samujjal Das dhingrak 676 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Its answer is A) 25, when all a,b,c,d and e are=1, in all other cases It will be greater than that. Try Hit and trial and option elimination Prateeksha Keshari answered Dec 9, 2014 Prateeksha Keshari comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes direct application of power mean inequality gives the answer as 25 Anand Vijayan answered Mar 1, 2017 Anand Vijayan comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 1. 25 Because the if we increase the value of a,b,c,d,e then value of part-1 of equation increases and if we decrease the value the part-2 will be increases so this function is minimum at a=b=c=d=e=1; abhi1997 answered Mar 2, 2017 abhi1997 comment Share Follow See all 0 reply Please log in or register to add a comment.