CAT Question Bank  Maxima & Minima

Method 1 :
AM ≥ GM
(p + q + r)/3 ≥ (pqr)^1/3  (1)
(1/p + 1/q + 1/r)/3 ≥ (1/pqr)^1/3  (2)
(1) * (2) > 1/9 (p + q + r)(1/p + 1/q + 1/r) ≥ 1
(p + q + r)(1/p + 1/q + 1/r) ≥ 9
So minimum value = 9Method 2 :
(p + q + r)(1/p + 1/q + 1/r)
= p/p + p/q + p/r + q/p + q/q + q/r + r/p + r/q + r/r
= 1 + 1 + 1 + (p/q + q/p) + (p/r + r/p) + (q/r + r/q)
= 3 + 2 + 2 + 2 = 9 [ @ChhabbisGyarah ]Method 3 :
(p + q + r)(1/p + 1/q + 1/r) = (p + q + r)(pq + qr + pr)/pqr is minimum when pqr is maximum.
pqr is maximum when p = q = r
(p + q + r)(1/p + 1/q + 1/r) = 3p * 3/p = 9

Maximum occurs when a^2 = b^2 = c^2 = d^2 = 25
=> a = b = c = d = +/ 5
Maximum value of a + b + c + d = 5 + 5 + 5 + 5 = 20

@rowdyrathore
min = 20
max = 20

@rowdyrathore min = 7
max = 73
cauchy method

Graph is as below

@rowdyrathore can u plz solve this?