Question

If a, b, c are three positive real numbers, then the minimum value of expression b+ca+a+cb+a+bcb+ca+a+cb+a+bcis

Answer 1

Answer:

11

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

For three positive real numbers a, b and c

AM = a+b+c3

GM = abc−−−√3

HM = 3abcab+bc+ca=31a+1b+1c

The AM, GM and HM of positive numbers satisfy the inequality

HM≤GM≤AM

where equality holds when all numbers are equal. (Not providing the proof)

So for a,b and c we have

a+b+c3≥abc−−−√3≥3abcab+bc+ca

For our problem, we require only the AM and HM. So,

a+b+c3≥3abcab+bc+ca

Cross multiplying, we get

(a+b+c)(ab+bc+caabc)≥9

(a+b+c)(1a+1b+1c)≥9

For minimum value

(a+b+c)(1a+1b+1c)=9

Finally to our problem, we have

b+ca+c+ab+a+bc

=b+c+a−aa+c+a+b−bb+a+b+c−cc

=(a+b+c)(1a+1b+1c)−3

Now for minimum value we can substitute (a+b+c)(1a+1b+1c) with the result we obtained earlier ie 9.

Hence, the minimum value of

b+ca+c+ab+a+bc=9−3=6(ans)