Question

Given `a > 0, b > 0, c > 0` calculate the minimum for `(a^x+b^x+c^x)/3` ?

Answer 1

`( a b c )^(1/3)`

Explanation:

`a^x` = exp(x * ln(a)) = `1 + x*ln(a) + (x²(ln(a))²)/2 + ...`

For small x, we need only the first two terms.
So we have

`lim x->0 " "((1+x*ln(a)+1+x*ln(b)+1+x*ln(c))/3)^(1/x)`

= `lim x->0 " "(1 + (ln(a)+ln(b)+ln(c))*x/3)^(1/x)`

= `lim x->0 " " (1 + k x)^(1/x)` , with k= `(ln(a)+ln(b)+ln(c))/3`

= exp(k) `" "` (Euler's limit)

= `( a b c )^(1/3)`

Answer 2

`root(3)(abc)`

Explanation:

Assuming `a > 0, b > 0, c > 0` and `x ge 0`

`(a^x+b^x+c^x)/3 ge root(3)(a^xb^xc^x) = (abc)^(x/3)` then

`((a^x+b^x+c^x)/3)^(1/x) ge( (abc)^(x/3))^(1/x) = root(3)(abc)`

NOTE

`((a^x+b^x+c^x)/3)^(1/x)` for `x in (0, oo)` has a minimum at

`lim_(x->0)((a^x+b^x+c^x)/3)^(1/x)`

because

`d/(dx)((a^x+b^x+c^x)/3)^(1/x) gt 0` for `x in (0,oo)`