Question

the sum of n postive integers is 2012. the sum of their squares is greater than 4024.then the sum of their cubes is greater than?

Answer 1

1+2+3.....n=n(n+1)/2 =2012
1²+2²+3²+...+n²=n(n+1)(2n+1)/6=4048
1³+2³+3³+....+n³=n²(n+1)²/4=426544

Answer 2

Thus always the sum of cubes is more than  8048.

As we are given n positive numbers,  and nothing is specified that they are consecutive or natural numbers...  we cannot assume that.

==========================
Let  the n integers all be equal to 1:    then n = 2012.  There are 2012 integers of value1.
 sum of squares of all the numbers = 2012 * 1² = 2012  < 4024.

Let the n integers all be equal to  2.  Then we have 1006 integers of value 2.
 sum of squares = 1006 * 2² = 4024.

Suppose we have  n-2 integers of value 2, and remaining two integers are 1  and 3.
Then we have n = 1006.
 Sum of squares = 1004 * 2^2 + 1^2 + 3^2 = 4026  >  4024

Instead of  1 and 3 as above,  if we choose 4 numbers as 1, 1, 5, 1, and remaining 1002 numbers as 2, 
         the sum of squares is :  1002 * 2^2  + 1+1+1+5^2  = 4036 > 4024

Thus as we choose a bigger number than 2, the sum of squares is always more than  4024.


now sum of cubes for all numbers to be 2,  is
           1006 * 2³ = 8048.

if we choose,  1004 numbers as 2, another 1 and another 3,  then their cubes sum to :  1004 * 2³ +  1³ + 3³ = 8060  > 8048

if we choose ,  1002 numbers as 2, the remaining as 1,1, 1, 5, then  their cubes sum to :    1002 * 2³ + 1+1+1+5³ = 8144  > 8048

Thus always the sum of cubes is more than  8048.

================================
if there are    n positive integers of sum = M, then the sum of their squares is minimum when  all the integers are equal and have value = M/n.

same way the sum of cubes of the numbers is minimum when all integers are equal and have value =  M/n.

This is the principle, which I have used.