(a) The diagram above contains 13 boxes. The numbers in the second and the twelfth boxes are respectively 175 and 70. Fill up the boxes with natural numbers such that
(i) the sum of all numbers in all 13 boxes is 2015,
(ii) the sum of the numbers in any three consecutive boxes is always the same.
      The solution must contain the steps how you arrive at the numbers.
(b) If x, y, z are real and unequal numbers, prove that $$2015x^{2}+2015y^{2}+6z^{2} > 2(2012xy+3yz+3zx)$$.

For part (a), since every three consecutive boxes sum to the same constant, the sequence of 13 boxes repeats in a pattern of period 3, letting the first box be a and the third be b (with the second fixed at 175, and the twelfth position, which falls in the same repeating slot as the second box mod 3, fixed at 70, giving $$b=70$$). Summing all 13 boxes in terms of a, b and 175 and setting the total to 2015 gives $$5a+4b+700=2015$$, and substituting $$b=70$$ gives $$5a=1035$$, so $$a=207$$. For part (b), applying AM-GM separately to the pairs $$2012x^2,2012y^2$$; $$3y^2,3z^2$$; and $$3x^2,3z^2$$ gives three inequalities that, when added together, directly yield $$2015x^2+2015y^2+6z^2 \ge 2(2012xy+3yz+3zx)$$, with strict inequality holding since x, y, z are unequal.