Determine the minimum value of a/(2b) + b/(4c) + c/(8a) where a,b,c are positive real numbers?

2 Answers
May 9, 2017

3/4

Explanation:

Making a = lambda_1 b and b=lambda_2 c we have

a/(2 b) + b/(4 c) + c/(8 a) =lambda_1/2 + 1/(8 lambda_1 lambda_2) + lambda_2/4

Now with f(lambda_1,lambda_2) = lambda_1/2 + 1/(8 lambda_1 lambda_2) + lambda_2/4

and determining the stationary points with

{((partial f)/(partial lambda_1)=1/2 - 1/(8 lambda_1^2 lambda_2)=0),((partial f)/(partial lambda_2)=1/4 - 1/(8 lambda_1 lambda_2^2)=0):}

and solving for lambda_1, lambda_2 we get the real values

lambda_1 = 1/2, lambda_2 = 1

Those values are relative to a local minimum for f(lambda_1,lambda_2) because at that point the hessian

grad^2 f =1/2 ((1,1),(1,4)) with characteristic polynomial

p(s)=3/4 - (5 s)/2 + s^2=(s-1/4 (5 - sqrt[13]))(s-1/4 (5 + sqrt[13]))

have two positive roots, qualifying the point as a minimum point.

The value attained for f at this point is 3/4

May 9, 2017

(a/(2b)+b/(4c)+c/(8a))_(min)=3/4.

Explanation:

We use the AMGM Inequality (AG) to solve this Problem.

The AG Property :

AA x,y,z in RR^+, (x+y+z)/3 ge root(3)(xyz).

Applying this to x=a/(2b), y=b/(4c), z=c/(8a)," all "gt0," we have,"

1/3{a/(2b)+b/(4c)+c/(8a)} ge root(3){a/(2b)*b/(4c)*c/(8a)}.

rArr 1/3{a/(2b)+b/(4c)+c/(8a)} ge root(3)(1/64)=1/4.

rArr a/(2b)+b/(4c)+c/(8a) ge 3/4.

:. (a/(2b)+b/(4c)+c/(8a))_(min)=3/4, as derved by Respected

Cesareo R., Sir!

Enjoy Maths.!