If a, b, c are positive such that #ab^2c^3 = 64# then the least value of #(1/a + 2/b + 3/c)# is?

Question

A) # 6#
B) #2#
C) #3#
D) #32#

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Answer 1 · 2017-08-24T13:15:09

I got C

Note that #64 = (2^3)(2^2)(2)#, but this would mean that #a = b = c# which is not stated in the question.

#a/2 + b/2 + c/2 = 1/2 + 2/2 + 3/2 = 2 +1 = 3#

Please let me know if this is the desired answer.

Hopefully this helps!

Answer 2 · 2017-08-24T15:35:09

See below.

Using inequalities

#1/n sum_(k=1)^n x_k ge (prod_(k=1)^n x_k)^(1/n)# now making

#x_k=cdots=x_j=u#, #m# times,
#x_(j+1)=cdots=x_xi = v#, #p# times,
#x_(xi+1)=cdots=x_n = w#, #q# times,

we get to a particular case

#1/(m+p+q)(m u+pv+qw) ge (u^m v^p w^q)^(1/(m+p+q))#

now making

#u=1/a, v=1/b,w=1/c# we have

#1/6(1/a+2/b+3/c) ge (1/(a b^2 c^3))^(1/6)= (1/64)^(1/6)=1/2# and finally

#1/a+2/b+3/c ge 3#