Question

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

Answer 1

`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`

Answer 2

`(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.!