SAMC

Assume, by contradiction, that $a>8$. Then $14-a<6$. Applying AM-GM inequality, we get $$6>14-a=a^x+a^y+a^z\ge 3a^{\frac{x+y+z}{3}}>3\cdot 8^{\frac{1}{3}}=3\cdot 8^{\frac{1}{3}}=6,$$ a contradiction.