n bag of sand and one algorithms We have $n$ bags of sand, with volume $$v_1,...,v_n, \forall i: \space 0 < v_i < 1$$ but not essentially sorted. we want to place all bag to boxes with volumes 1. We propose one algorithm:

At first we place all bags in the original order. Then we select one
  box and place on it, bag $1, 2, 3,...$ until these can be place in box.
  If the $i^{th}$ bag couldn't be inserted in a box, we choose another box and
  place it $i^{th}, i^{th+1},...$ until these can be place in the box.

If the number of boxes that be used in this algorithms $X$, and the number of boxes used in minimum way (by using minimum algorithms) be $Y$, why always $X > Y$ is false, and always true that $X < 2Y$.
 A: If $Y$ is the minimum number of boxes used by any algorithm and $X$ is the number of boxes used by your algorithm, then it must be that $X\ge Y$ simply by the way we defined $Y$.
If $X<Y$, then $Y$ couldn't be the minimum number of boxes used by any algorithm because your algorithm produces a smaller result.
It may be that you can prove that $X<2Y$, but that doesn't contradict anything about our discussion above.
A: In the best efficient way, each box is completely full:
$$Y=\Sigma v_i$$
So in general
$$Y\ge \Sigma v_i$$
In addition, in your algorithm, each two consecutive box must have at least one unit sand (otherwise the first box could accommodate the next one). So, even in the worst case:
$$\Sigma v_i>\frac X2$$
when X is even and
$$\Sigma v_i>\frac {X-1}2$$
when X is odd.
Combining two in-equations:
$$\frac X2 < \Sigma v_i \le Y$$
or
$$X < 2Y$$
for when X is even.
for odd cases of X we have 
$$X < 2Y+1$$
However, since 2Y is even, X=2Y does not happen and it can be re-written as:
$$X < 2Y$$
So, for both odd and even cases of X:
$$X < 2Y$$
