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Sum Square Roots ≤ Real Number (Posted on 2010-04-10) Difficulty: 3 of 5
Determine all possible value(s) of a positive integer constant c that satisfy this relationship:

√(y-1) + √(y-2) + .....+ √(y-c) ≤ y

whenever y is a positive real number ≥ c

No Solution Yet Submitted by K Sengupta    
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analytical solution | Comment 3 of 4 |
let f(c,y) = sum( sqrt(y-t), t=1 to c)
then we want to know for what integers c does
y>=c imply f(c,y)<=y
now f(c,c)>=integral sqrt(x) x=0 to c-1
f(c,c) >= (2/3)*(c-1)^(3/2)
now we want to know when (2/3)*(c-1)^(3/2)<=c
4*(c-1)^3<=9c^2
4c^3-4c^2+4c-4<=9c^2
4c^3-13c^2+4c-4<=0
which implies c<=3
so now we need to check if c=1,2 or 3 work

c=1
this is simply sqrt(y-1)<=y when y>=1 which is true

c=2
sqrt(y-1)+sqrt(y-2)<=y
y-1+2*sqrt((y-1)(y-2))+y-2<=y^2
2*sqrt((y-1)(y-2))<=y^2-2y=y(y-2)
4*(y-1)(y-2)<=y^2(y-2)^2
4*(y-1)<=y^2(y-2)
4y-1<=y^3-2y^2
y^3-2y^2-4y+1>=0
the polynomial has no real solutions and thus never crosses the x-axis since it is positive for y=0 then it is positive for all y thus it works for c=2

c=3
sqrt(y-1)+sqrt(y-2)+sqrt(y-3)<=y
looking at y=4 we get
sqrt(3)+sqrt(2)+1<=4
which is not true, thus it does not hold for c=3


thus it only holds for c=1 and c=2

Edited on April 11, 2010, 7:08 am
  Posted by Daniel on 2010-04-10 19:01:31

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