answer:1. Given the inequalities: [ sum_{i leq j} a_i a_j leq 1 quad text{and} quad sum_{i leq j} b_i b_j leq 1 ] 2. We need to prove that: [ sum_{i leq j} left( a_i - b_i right) left( a_j - b_j right) leq 1 ] 3. Consider the expression (left( a_i - b_i right) left( a_j - b_j right)): [ left( a_i - b_i right) left( a_j - b_j right) = a_i a_j + b_i b_j - left( a_i b_j + a_j b_i right) ] 4. Summing over all ( i leq j ): [ S = sum_{i leq j} left( a_i - b_i right) left( a_j - b_j right) = sum_{i leq j} a_i a_j + sum_{i leq j} b_i b_j - sum_{i leq j} left( a_i b_j + a_j b_i right) ] 5. Notice that the sum (sum_{i leq j} left( a_i b_j + a_j b_i right)) can be bounded as follows: [ sum_{i leq j} left( a_i b_j + a_j b_i right) geq left( a_1 + a_2 + a_3 right) left( b_1 + b_2 + b_3 right) ] 6. Let's assume, without loss of generality, that: [ a_1 + a_2 + a_3 geq b_1 + b_2 + b_3 ] Therefore: [ sum_{i leq j} left( a_i b_j + a_j b_i right) geq left( b_1 + b_2 + b_3 right)^2 ] 7. Then, substituting into the inequality for (S), we get: [ S leq sum_{i leq j} a_i a_j + sum_{i leq j} b_i b_j - left( b_1 + b_2 + b_3 right)^2 ] 8. Using the given inequalities, we find: [ S leq sum_{i leq j} a_i a_j + sum_{i leq j} b_i b_j leq 1 ] 9. Also, given: [ left( b_1 + b_2 + b_3 right)^2 geq sum_{i leq j} b_i b_j ] It is clear that: [ left( b_1 + b_2 + b_3 right)^2 geq sum_{i leq j} b_i b_j ] 10. Combining all these results we can conclude: [ sum_{i leq j} left( a_i - b_i right) left( a_j - b_j right) leq 1 ] Therefore, the proof is complete. (blacksquare)

answer:For the function `sin x`, which oscillates between -1 and 1, all potential solutions for frac{x}{50} = sin x must lie in the interval `[-50, 50]` because beyond this range, `frac{x}{50}` would exceed the maximum and minimum values that `sin x` can attain. The equation can be visualized graphically. The function `y = frac{x}{50}` is a straight line through the origin with a slope of `frac{1}{50}`, and `y = sin x` is a periodic function. The point where the sine function reaches 1 (i.e., its maximum) at `x = (2n + frac{1}{2})pi` for `n` being an integer should be below 50. To find the last '`n`', consider: [ (2n + frac{1}{2})pi < 50 ] Solve for `n` to find the largest integer within the admitted range. This rough calculation indicates that: [ 2n + frac{1}{2} < frac{50}{pi} approx 15.92 ] Hence, `n` can be at most 7 (using integer part, because 15.5 - 0.5 = 15). Within each period `2pi`', `y = sin x` and `y = frac{x}{50}` intersect twice (once ascendant and once descendant). Between `-50` and `50`, there are eight full `2pi` periods (from `-4pi` to `4pi`). Thus, we have: [ 2 times 8 = 16 ] solutions in the interval `[-4pi, 4pi]`. Together with the end periods, which need closer verification but one would conclude that they each add two solutions as they do not exceed the `[-50, 50]` limit, we can confirm the total number of solutions. The symmetry of sine curves around the origin leads to double counting at `x=0`. Thus, the total number of solutions is: [boxed{33}] (16 from main intervals, and counting slack periods as fully available).

answer:Define T = sum_{n=0}^infty frac{G_n}{5^n} = sum_{n=0}^infty frac{2F_n}{5^n} = 2 sum_{n=0}^infty frac{F_n}{5^n}. Let S' = sum_{n=0}^infty frac{F_n}{5^n}. Then using the method from the original problem: [ S' = F_0 + frac{F_1}{5} + frac{F_2}{5^2} + frac{F_3}{5^3} + dots = frac{F_0 + 1}{5} + frac{F_1}{5^2} + frac{F_2}{5^3} + frac{F_3}{5^4} + dots ] [ S' = frac{1}{5} + frac{F_0}{5^2} + frac{F_1}{5^3} + frac{F_2}{5^4} + dots + frac{1}{5} S' + frac{1}{5^2} S' ] [ S' = frac{1}{5} + frac{1}{5} S' + frac{1}{5^2} S' ] [ S' left( 1 - frac{1}{5} - frac{1}{25} right) = frac{1}{5} ] [ S' = frac{frac{1}{5}}{1 - frac{6}{25}} = frac{frac{1}{5}}{frac{19}{25}} = frac{5}{19} ] So, T = 2S' = 2 cdot frac{5}{19} = frac{10}{19}. Conclusion: [ boxed{T = frac{10}{19}} ]

answer:To solve this problem, we follow a systematic approach: # Part 1: Find the Expression of the Quadratic Function Given that the vertex of the quadratic function is at left(2,-4right), we can express the quadratic function in vertex form as: [y = a(x - 2)^2 - 4] We are also given that the quadratic function intersects the y-axis at y = 4. The x-coordinate for any point on the y-axis is 0, so we substitute left(0, 4right) into our equation to find a: [ begin{align*} 4 &= a(0 - 2)^2 - 4 4 &= 4a - 4 8 &= 4a a &= 2 end{align*} ] Therefore, the expression of the quadratic function is: [y = 2(x - 2)^2 - 4] Encapsulating the final answer for part (1): [boxed{y = 2(x - 2)^2 - 4}] # Part 2: Check if Point P(3,5) is on the Graph To determine if the point P(3,5) lies on the graph of our quadratic function, we substitute x = 3 into the equation and check if the output is y = 5: [ begin{align*} y &= 2(3 - 2)^2 - 4 &= 2(1)^2 - 4 &= 2 - 4 &= -2 end{align*} ] Comparing the calculated y-value to the y-coordinate of point P, we find: [5 neq -2] Since the calculated y-value does not match the y-coordinate of point P, we conclude that point P(3,5) is not on the graph of this quadratic function. Encapsulating the final answer for part (2): [boxed{text{No, the point } P text{ is not on the graph of this quadratic function.}}]