Inequality of arithmetic mean of two sets If $a,b>0$ and $Q=\{x_1, x_2, x_3,..., x_a\}$ a subset of the natural numbers $1, 2, 3,..., b$ such that, for $x_i+x_j<b+1$ with $1 ≤ i ≤ j ≤ a$, then $x_i+x_j$ is also an element of Q. Prove that: 
$ \frac{x_1+x_2+x_3+...+x_a}{a} ≥ \frac{b+1}{2}$ so basically, you have to prove that the arithmetic mean of Q satisfying the condition is greater than equal to the arithmetic mean of the set the natural numbers $1, 2, 3,..., b$.
Any help would be appreciated. Thanks in advance!

Hint: For each $i$, show that $ x_i + x_{a-i}  \geq b+1$.

 Proof by contradiction. Suppose $x_i + x_{a-i} < b+1$, which satisfies the condition in the problem.
 What does it make sense to consider next?
 Do some work to reach a contradiction.

Corollary: $ 2\sum x_i \geq a (b+1)$, and the result follows.
