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!

I think i have solved my problem. @Calvin Lin thanks a lot, the hint was very helpful. Please correct me if I am wrong. Solution:  Without loss of generality we assume $x_1<x_2 < ....<x_{a−1}< x_a$ $(1)$. Suppose $x_i+x_{a−i}<b+1 $.  Because of $(1)$ we know that: $x_i+x_{a−i} <b+1 \Rightarrow x_i+x_{a−i-1} <b+1\Rightarrow x_i+x_{a−i-2} <b+1 \Rightarrow ...\Rightarrow x_i+x_{2}<b+1\Rightarrow x_i+x_{1}<b+1\Rightarrow x_i<b+1\Rightarrow x_{i-1}+x_{1}<b+1\Rightarrow x_{i-2}+x_{1}<b+1\Rightarrow ...\Rightarrow x_{2}+x_{1}<b+1\Rightarrow x_{1}+x_{1}<b+1\Rightarrow x_{1}<b+1.$ Thats why Q has to have $a-i+i+1=a+1$ elements. Contradiction! Thats why for each i,  $x_i+x_{a−i}≥b+1$ must hold. After that the proof is excatly like Calvin Lin described.
