Proof verification of a problem regarding positivity of sum of 10 real numbers Question:

We are given ten real numbers such that the sum of any four of them is
positive. Show that the sum of all ten numbers is also positive.

My approach:Let the real numbers be $a_1,a_2,.....,a_{10}$. Let $S$ denote the sum of all ten of those real numbers.
Now,let us assume that $S \leq 0$. If so then,
$$S=\sum_{i=1}^{10} a_i \leq 0 \Rightarrow \sum_{i=3}^{10} a_i +a_1+a_2 \leq 0$$
If $a_1+a_2\geq 0$ then we get that $S>0$ contrary to our assumption. Thus, $a_1+a_2 < 0$
Thus,we see that,  $$ S_1=\sum_{i=3}^{10} a_i +a_1+a_2+a_1+a_2<a_1+a_2<0$$
But,then we see that,
$$S_1=\sum_{i=1}^{8} a_i +a_9+a_{10}+a_1+a_2>0$$ by using the hypothesis.Thus, we arrive at a contradiction. Hence, we can conclude that $S$ must be positive.
My request:Please let me know if there is a glitch in the proof. Further,Help me find a shorter and more elegant proof to this problem. An elegant proof will be much appreciated.
 A: Assume, without loss of generality, that $a_1 \leq a_2 \leq \cdots \leq a_{10}$. Since $\sum_{i=1}^4 a_i > 0$ and $a_i >0$ for $i\ge 4$ (otherwise the first sum could not be positive), the result follows.
A: Consider $(a_1+a_2+a_3+a_4)+(a_1+a_2+a_3+a_5)+(a_1+a_2+a_3+a_6)+\dots + (a_7+a_8+a_9+a_{10})$
That is, $\sum\limits_{\Delta\in \binom{[10]}{4}}\sum\limits_{i\in \Delta}a_i$, be the sum of all sums of four terms.
This can be seen to be equal to $\binom{9}{3}(a_1+a_2+a_3+a_4+\dots+a_{10})$ and is clearly positive as every parenthetical phrase in the sum is positive by hypothesis.

In case this is not clear, write the sum I consider as the following:
$\begin{array}{ccccccc}&a_1&+a_2&+a_3&+a_4\\+&a_1&+a_2&+a_3&&+a_5\\+&a_1&+a_2&+a_3&&&+a_6\\\vdots\\+&&a_2&+a_3&+a_4&+a_5\\\vdots\end{array}$
If we look at the sum by first adding left-to-right followed by top-to-down, these are each sums of four of your numbers and positive by hypothesis so the overall sum is the sum of strictly positive numbers and thus positive.
If we were to look at the sum instead by adding top-to-down followed by left-to-right, we see that each number occurs in precisely $\binom{9}{3}$ of the rows by elementary counting principles and so the overall sum is $\binom{9}{3}(a_1+a_2+\dots+a_{10})$.
As the two methods of arriving at the sum must be equal we see that $\binom{9}{3}(a_1+a_2+\dots+a_{10})$ must be positive.  Dividing by $\binom{9}{3}$, also a positive number, does not change the sign, so your sum of all ten numbers must also be positive.
A: Let $ a_{11} = a_1, a_{12} = a_2, a_{13}  = a_3$. Then,
$$ 0 <  \sum_{i=1}^{10} a_i + a_{i+1} + a_{i+2} + a_{i+3}  = 4 \sum_{i=1}^{10} a_i$$
