Guest

if a,b,c are positive real numbers such that loga/(b-c) = logb/(c-a) = logc/(a-b) then proove that a (b-c) + b ( c-a) + c (a-b) >= 3 a (a) + b (b ) + c (c) >= 3

 
if a,b,c are positive real numbers such that 
loga/(b-c) = logb/(c-a) = logc/(a-b)
then proove that 
  1. a(b-c)+ b(c-a) + c(a-b) >= 3
  2. a(a)+ b(b+ c(c) >= 3

Grade:11

1 Answers

Samyak Jain
333 Points
5 years ago
Let loga / (b – c) = logb / (c – a) = logc / (a – b) = k
\therefore loga = k (b – c) ,  a = 10 k (b – c)
    logb = k (c – a) , b = 10 k (c – a)
    logc = k (a – b) , c = 10 k (a – b)
 a(b – c) = {10 k (b – c)}^(b – c)  = 10 k (b – c)^2
b(c – a) = {10 k (c – a)}^(c – a)  = 10 k (c – a)^2
c(a – b) = {10 k (a – b)}^(a – b)  = 10 k (a – b)^2
 a(b – c) b(c – a) + c(a – b)  = 10 k (b – c)^2 10 k (c – a)^2 10 k (a – b)^2             ….........(1)
From (1), the minimum value of a(b – c) b(c – a) + c(a – b) occurs when a = b = c which is 3 (you can check).
\therefore  a(b – c) b(c – a) + c(a – b)  \geq  3
a(a)+ b(b+ c(c)  = 10 k (b – c) a 10 k (c – a) b + 10 k (a – b) c                                          ….............(2)
From (2), the minimum value of a(a)+ b(b+ c(c)  occurs when a = b = c again which is 3.
\therefore a(a)+ b(b+ c(c)  \geq  3

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free