Assume that we have six positive real numbers whose sum is 150. Prove that there exist two of them whose difference is less than 10.

Solution 1:

Your argument fails because some of the parenthesized terms must be negative. Based on the symmetry you can renumber things so that $a_1 \gt a_2 \gt a_3 \gt a_4 \gt a_5 \gt a_6$ You know $a_6 \gt 0$ so $a_5 \gt 10$. Then $a_4 \gt 20$. Keep going.

Solution 2:

Best to start with the smallest one $x>0 $. Then the others in increasing order are $$x+d_1, x+d_1 + d_2, \ldots,x+ d_1 + d_2+d_3 + d_4 + d_5$$ where $d_i\ge 0$ is the difference between $i+1$ th and $i$th number. Now their sum is $$150 = 6 x + 5 d_1 + 4 d_2 + 3 d_3 + 2 d_4 + d_5$$ so if $d$ is the smallest of the $d_i$ we get $$150 > (5+4+3+2+1) d = 15 d$$ so $$0\le d < 10$$ In general, if $n$ positive numbers have sum $S$ then two of them have difference $\ge 0$ and $< S/\binom{n}{2}$

Solution 3:

Building on @fleablood's answer: the minimum number may be arbitrarily small, but still positive. We call it a.

The next five numbers (at their smallest) must be:

$$b = 10 + a$$ $$c = 20 + a$$ $$d = 30 + a$$ $$e = 40 + a$$ $$f = 50 + a$$ Thus, the sum of all the numbers is

$$a + b + c + d + e + f = 150 + 6a$$

Since a is positive (though arbitrarily small) the sum must be greater than 150, which is a contradiction.

Solution 4:

The smallest of the six terms must be more than $0$.

If the differences of any two is at least $10$, then the second must be more than $10$

The smallest the third must be more than $20$.

Etc.

So all six must add up to more than $0+10+20+30+40+50=150$.

So they can't add to $150$. They can be arbitrarily close to $150$ but they must be more than $50$.

.....

If the smallest one is $a = \epsilon > 0$ and then the next smallest is $b \ge 10 + \epsilon$ and then the next smallest is $c \ge 10 + b \ge 20 +\epsilon$. Etc.

The sum is $a + b + c + d+e +f \ge \epsilon + (10 + \epsilon) + .... + (50 + \epsilon) = 150 + 6\epsilon > 150$.