answer:By applying the Cauchy-Schwarz inequality, we have [ [(a+b) + (a+c) + (a+d) + (b+c) + (b+d) + (c+d)] left( frac{1}{a+b} + frac{1}{a+c} + frac{1}{a+d} + frac{1}{b+c} + frac{1}{b+d} + frac{1}{c+d} right) ge (1+1+1+1+1+1)^2 = 36. ] Given that (a + b + c + d = 2), we observe that [ (a+b) + (a+c) + (a+d) + (b+c) + (b+d) + (c+d) = 3(a+b+c+d) = 6. ] Thus, substituting this into the inequality, [ frac{1}{a+b} + frac{1}{a+c} + frac{1}{a+d} + frac{1}{b+c} + frac{1}{b+d} + frac{1}{c+d} ge frac{36}{6} = 6. ] Equality holds if (a = b = c = d = frac{1}{2}). Conclusion: The minimum value of the expression is (boxed{6}).

answer:We have 36 possible outcomes when rolling a die twice, which are listed below: (1, 1), (1, 2), dots , (1, 6), (2, 1), (2, 2), dots , (2, 6), quad vdots (6, 1), (6, 2), dots , (6, 6). In order for three line segments of lengths a, b, and 5 to form an isosceles triangle, it must be true that a+b>5. The possible outcomes satisfying this condition are: (1, 5), (5, 1), (2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4), (5, 5), (6, 5), (5, 6), (3, 3), (4, 4), (6, 6). There are 14 such outcomes. Therefore, the probability P that three line segments of lengths a, b, and 5 can form an isosceles triangle is given by P=frac{text{number of favorable outcomes}}{text{total number of outcomes}}=frac{14}{36}=boxed{frac{7}{18}}.

answer:We are given a game where two players take turns removing between 1 and 9 chips from a pile of 110 chips, with the restriction that a player cannot repeat their opponent's move. The player who cannot make a move loses the game. We need to determine which player has a winning strategy. Step 1: Determine Winning Positions We construct a table where each row corresponds to the number of chips left, and the columns indicate the winning moves. | Chips Number | Winning Moves | | :---: | :--- | | 0 | - | | 1 | 1 | | 2 | 1, 2 | | 3 | 3 | | 4 | 4 | | 5 | 5 | | 6 | 3, 6 | | 7 | 7 | | 8 | 4, 8 | | 9 | 9 | | 10 | 5 | | 11 | - | | 12 | 1 | | 13 | 1, 2 | | 14 | 3, 7 | | 15 | 4, 5 | | 16 | 5 | | 17 | 6 | | 18 | 7, 9 | | 19 | 8 | | 20 | 9 | | 21 | 5 | | 22 | - | | 23 | 1, 6 | | 24 | 2 | | 25 | 3 | | 26 | 2, 4, 5 | | 27 | 5, 8 | | 28 | 3, 6 | | 29 | 7, 9 | | 30 | 8 | | 31 | 9 | | 32 | - | We extend this table as much as necessary, noting the pattern that we look for positions where players cannot make a winning move. Step 2: Identify Patterns and Winning Strategy The table shows that certain positions (e.g., 0, 11, 22, 32) are losing positions when no matter what the player does, the opponent can always force a win. Step 3: Determine the Winning Strategy from Starting Position Our initial pile has 110 chips. We must determine if starting with 110 chips fits into one of these patterns. The periodicity observed in the table suggests that every 32 chips, the game state repeats itself. Notice: [ 110 equiv 14 pmod{32} ] This means, after taking some chips, the game state of 110 chips is similar to the game state of 14 chips. From the table, we see that with 14 chips (since 14 is (3) away from a losing position (11)), one player can always force the game into a losing state for the opponent. Step 4: Calculate Specific Winning Moves We verify positions with 110 chips based on modulo results: [ 110 - 32 times 3 = 14 ] From the number of chips: 1. Removing 3 chips will lead to (107) chips. 2. Removing (7) chips will lead to (103) chips. With these subtractions, (110) chips reduces to (14), making it a strategic position for the beginner to win since (14) is a losing position as observed from our list. # Conclusion: Given the initial number of 110 chips, the first player has a winning strategy based on the modulo (32) periodicity and positions available from the generated table. Therefore, the first player can guarantee a win by forcing the second player into a losing position. [boxed{text{The first player has a winning strategy.}}]

answer:To represent the situation with a system of equations, we can use the following information: 1. The total number of carrots eaten from Tuesday to Thursday is 15. 2. Wilfred eats twice as many carrots on Thursday as he did on Wednesday. From this information, we can set up the following equations: Equation 1 (Total carrots eaten): x + y + z = 15 Equation 2 (Twice as many carrots on Thursday as on Wednesday): z = 2y Now we have two equations with three variables. To solve for x, y, and z, we need another equation or we can express one variable in terms of the others. Since we have z in terms of y, we can substitute z in the first equation: x + y + 2y = 15 x + 3y = 15 Now we have two equations with two variables (x and y): Equation 1 (modified): x + 3y = 15 Equation 2 (as before): z = 2y We still need one more equation to solve for all three variables. However, we can choose a value for y and then solve for x and z. Since the problem does not give us any more constraints, we can choose a value for y that makes sense. For example, let's say y = 3 (Wilfred eats 3 carrots on Wednesday). Then we can solve for x and z: x + 3(3) = 15 x + 9 = 15 x = 15 - 9 x = 6 Now we can find z using the second equation: z = 2y z = 2(3) z = 6 So, if Wilfred eats 3 carrots on Wednesday, he will eat 6 carrots on Tuesday and 6 carrots on Thursday to reach a total of 15 carrots over the three days. However, this is just one solution. There are infinitely many solutions to this problem since we can choose different values for y. For example, if y = boxed{2,} then x would be 9 and z would be 4, and so on. The problem does not specify any additional constraints, so any combination of x, y, and z that satisfies the two equations is a valid solution.