Question

Suppose you have 5 real numbers whose sum
of squares is equal to 5. What is the maximum
value of the sum of cubes of these 5 numbers?

Please show step

Answer 1

You don't say what kind of numbers they are ,
e.g. if they are rationals
a=1
b = .5
c= 1.1
d= 1.4
sum of squares of those 4 = 4.42
then the 5th number is √.58

if they are integers, they must be
±1,±1,±1,±1,±1 or ±2,±1,0,0,0

e.g. 2^2 + 1^2 + 0^2 + 0^2 = 5

The sum of cubes of the first set is 5
the sum of cubes of the second set is 9

so the numbers must be 2, 1, 0, 0, 0


incidentally the sum of cubes of my first example would be
5.6417...

On the other hand, suppose the 5 numbers are
<b>√5, 0, 0, 0, 0</b>
the sum of their squares is indeed 5
and the sum of their cubes is (√5)^3 or 11
18...

how about that?

Was there more information?

Answer 2

and the sum of their cubes is (√5)^3 or 11.18...