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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

2 Answers

Vikas TU
14149 Points
7 years ago
loga/(b-c) = logb/(c-a) = logc/(a-b) = k (let)(b-c)(loga)/(b-c)^2 = ka^(b-c) = e^(k(b-c)^2)......................(1)Similarly,c^(a-b) = e^(k(a-b)^2)....................(2)b^(c-a) = e^(k(c-a)^2)......................(3)Adding all the eqns. a(b-c)+ b(c-a) + c(a-b) = e^(k(b-c)^2) + e^(k(a-b)^2) + e^(k(c-a)^2) >=3as a,b,c > 0 and are positive real numbers.chec for minimumvalue of a,b,c.i.e a,b,cfor a = b =c =1 it gives 3 minum and for any other no. greater than 1 it will be greater than 3.
Sukalpa Mishra
23 Points
7 years ago
vikas yadav i see you have tried to answer the previous question also posted by me ...but that i had posted wrongly ther eal question is this and it has ab+c not ab-c as you might have thought from your previous questions..thanks in adavance if you are able to answer this question

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