answer:1. Let b_i = frac{a_i}{i} for i=1, 2, dots. We aim to prove that [ b_1 + b_2 + cdots + b_n geq a_n quad text{for all } n=1, 2, dots ] by induction on n. 2. **Base Case**: For n=1, we have b_1 = frac{a_1}{1} = a_1, so [ b_1 geq a_1 Rightarrow a_1 geq a_1. ] The base case holds trivially. 3. **Induction Hypothesis**: Assume that for some integer k geq 1, we have [ b_1 + b_2 + cdots + b_k geq a_k. ] 4. **Induction Step**: We must show that [ b_1 + b_2 + cdots + b_n geq a_n. ] Notice that [ b_1 + b_2 + cdots + b_n = frac{a_1}{1} + frac{a_2}{2} + cdots + frac{a_n}{n}. ] We need to prove that [ frac{a_1}{1} + frac{a_2}{2} + cdots + frac{a_n}{n} geq a_n. ] 5. Recall the definition b_i = frac{a_i}{i}, our goal can be written as [ b_1 + b_2 + cdots + b_{n-1} + b_n geq a_n. ] 6. We need to show [ n b_1 + (n-1) b_2 + cdots + b_n geq n a_n. ] This translates into: [ n frac{a_1}{1} + (n-1) frac{a_2}{2} + cdots + frac{a_n}{n} geq n a_n. ] 7. Next, we simplify this expression by combining terms: [ n b_1 + 2(n-1) b_2 + cdots + n b_{n-1} = (n-1) b_1 + (n-2) b_2 + cdots + b_{n-1} + b_1 + 2 b_2 + cdots + (n-1) b_{n-1}. ] 8. Rewriting terms, we have: [ begin{aligned} n b_1 + & cdots + n b_{n-1} = (n-1) b_1 + (n-2) b_2 + cdots + b_{n-1} + b_1 + 2 b_2 + cdots + (n-1) b_{n-1} & = b_1 + left(b_1 + b_2right) + cdots + left(b_1 + b_2 + cdots + b_{n-1}right) + left(a_1 + a_2 + cdots + a_{n-1}right). end{aligned} ] 9. From the initial inequality a_{i+j} leq a_i + a_j, we can see that [ 2 left(a_1 + a_2 + cdots + a_{n-1}right) geq (n-1) a_n. ] Therefore, [ sum_{i=1}^{n-1} left(a_i + a_{n-1}right) geq (n-1) a_n. ] 10. Thus we establish that: [ boxed{a_1 + frac{a_2}{2} + frac{a_3}{3} + cdots + frac{a_n}{n} geq a_n quad text{for all } n in mathbb{N}}. ]

answer:(1) Given that b cdot cos A + a cdot cos B = -2c cdot cos C, By the sine law, we have sin B cdot cos A + sin A cdot cos B = -2 sin C cdot cos C, This implies sin(A + B) = -2 sin C cdot cos C, Since A, B, and C are angles of a triangle, we know that sin(A + B) = sin C neq 0, Hence, cos C = -frac{1}{2}, Thus, C = frac{2pi}{3}. (2) Given the area of triangle ABC, we have frac{1}{2}ab sin C = 2sqrt{3}, Substituting b = 2a and sin C = frac{sqrt{3}}{2}, we get frac{1}{2} cdot a cdot 2a cdot frac{sqrt{3}}{2} = 2sqrt{3}, This simplifies to a = 2, So, b = 4, Now, using the cosine law, we have c^2 = a^2 + b^2 - 2ab cos C, Substitute a = 2, b = 4, and cos C = -frac{1}{2} into the equation, we get c^2 = 4 + 16 - 2 cdot 2 cdot 4 cdot (-frac{1}{2}) = 28, Hence, c = boxed{2sqrt{7}}.

answer:Sally and Fred gave Sara 4 onions, and they have 10 onions left. To find out the total number of onions they had before giving any to Sara, we add the 4 onions given to Sara to the 10 onions they have left: 4 onions (given to Sara) + 10 onions (left) = 14 onions (total before giving any away) We know that Sally grew 5 onions, so to find out how many onions Fred grew, we subtract the number of onions Sally grew from the total number of onions they had before giving any away: 14 onions (total) - 5 onions (Sally's onions) = 9 onions Therefore, Fred grew boxed{9} onions.

answer:Ned could buy 5 candies, and each candy costs 9 tickets. So, the total cost of the candies is: 5 candies * 9 tickets/candy = 45 tickets Ned won 19 tickets playing 'skee ball'. To find out how many tickets he won playing 'whack a mole', we subtract the tickets won from 'skee ball' from the total tickets needed to buy the candies: 45 tickets (total needed for candies) - 19 tickets (won from skee ball) = 26 tickets Therefore, Ned won boxed{26} tickets playing 'whack a mole'.