ISI question

[RichaS]

[Abhitesh]

Use AM >= GM
(Answer) >=3

[Nikkita]

How using AM>= GM method??

Yes, I am getting same answer by taking some different types of examples.. Not any formal method..

[Abhitesh]

[x1/x2 + x2/x3 +x3/x1]/3 >= (x1/x2 * x2/x3 * x3/x1)^1/3.

This can also be solved using calculus. Assume x1/x2 = x and x2/x3 = y and then minimize the function.
What's your method?

[Nikkita]

Note that Lowest No.=3 When x1,x2,x3= 1

Explanation:
To cover cases of integers as well as fractions, critical point is 1.
So, Since all nos. are positive,
Case 1) All x1,x2,x3< 1, (say all= 1/2) then [x1/x2 + x2/x3 +x3/x1]>=3
Case 2) All x1,x2,x3>1, (say all=2) then [x1/x2 + x2/x3 +x3/x1]>=3
Case 3) All x1,x2,x3=1, then [x1/x2 + x2/x3 +x3/x1]=3
 
Can take other 1-2 cases too where one or two of them>1, other < 1; everytime [x1/x2 + x2/x3 +x3/x1]>=3 definitely.

[Eyes-on-nobelP]

@Raghuram Can you elaborate on this calculus method suggested by you?

[Abhitesh]

With above substitution the function becomes
f(x,y) = x + y + 1/xy
First order condition
fx = 1 - 1/x^2y = 0
fy = 1 - 1/y^2x = 0
Solve these to get x=y=1.
Second order condition
fxx*fyy - fxy^2 > 0 for minimum.

Note - The generalised second order condition should also have fxx>0. This comes from the concept of Hessian.