This question is from AMTI 2016 .So, solve this problem as fast as you can.

If a, b, c are positive real numbers.Then find the minimum value of

(a+3c/a+2b+c) + (4b/a+b+2c) - (8c/a+b+3c)

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Solution

If a = b= c then we get
(a+3c/a+2b+c) + (4b/a+b+2c) - (8c/a+b+3c)
= 4a/4a+ 4a/4a - 8a/4a
= 1+1 -2 = 0
So minimum value = 0 when a=b=c

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