B Minimum value

1. Feb 28, 2017

matrixone

if $a,b,c,d,e$ are positive real numbers, minimum value of $$(a+b+c+d+e)( \frac{1}{a} +\frac{1}{b} +\frac{1}{c} +\frac{1}{d} +\frac{1}{e} )$$
(A) 25
(B) 5
(C) 125
(D) cannot be determined

My approach :
expanding the expression , i get
$$5+a( \frac{1}{b} +\frac{1}{c} +\frac{1}{d} +\frac{1}{e} )+ similar.terms.of.b,c,d,e$$

I cant find any ways to make those expressions vanish .....and some hit and trial gives me ans as 25
but i also cant find any way to make them 20 .... :( ..
or is the answer cannot be determined ?
pls help ....i dont have the answer ...

2. Feb 28, 2017

Fightfish

Have you heard of the power mean inequalities? The hint is to make use of the fact that the arithmetic mean is always greater than or equal to the harmonic mean.
(and yes I'm not giving any explicit expressions here because I want you look them up and try to understand them - the Art of Problem Solving wiki is a good place to start)

3. Mar 1, 2017

matrixone

thanx a lot for the reference Sir ..... So, the direct application of power mean inequality (by putting k1=1 and k2= -1 )gives me the expression is greater than or equal to 25 .... and i hope that is the answer :)

Last edited: Mar 1, 2017