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question:Given that (a_{n} geqslant a_{n1} geqslant cdots geqslant a_{1} geqslant 0) and (sum_{i=1}^{n} a_{i}=1), prove that for any nonnegative real numbers ( x_{i} ) and ( y_{i} ) (where ( 1 leqslant i leqslant n )), the following holds: begin{array}{l} left(sum_{i=1}^{n} a_{i} x_{i}prod_{i=1}^{n} x_{i}^{a_{i}}right)left(sum_{i=1}^{n} a_{i} y_{i}prod_{i=1}^{n} y_{i}^{a_{i}}right) leqslant a_{n}^{2}left(n sqrt{sum_{i=1}^{n} x_{i} sum_{i=1}^{n} y_{i}}sum_{i=1}^{n} sqrt{x_{i}} sum_{i=1}^{n} sqrt{y_{i}}right)^{2} . end{array}
answer:# Problem Statement: Let a_n geqslant a_{n-1} geqslant cdots geqslant a_1 geqslant 0, where sum_{i=1}^n a_i = 1. Show that for any non-negative real numbers x_i and y_i (1 leq i leq n), we have [ left( sum_{i=1}^n a_i x_i - prod_{i=1}^n x_i^{a_i} right) left( sum_{i=1}^n a_i y_i - prod_{i=1}^n y_i^{a_i} right) leqslant a_n^2 left(n sqrt{sum_{i=1}^n x_i sum_{i=1}^n y_i} - sum_{i=1}^n sqrt{x_i} sum_{i=1}^n sqrt{y_i} right)^2. ] Step 1: Initial Step First, demonstrate the inequality: [ sum_{i=1}^n a_i x_i - prod_{i=1}^n x_i^{a_i} leqslant a_n left( n sum_{i=1}^n x_i - left( sum_{i=1}^n sqrt{x_i} right)^2 right). ] Step 2: Product Inequality By combining the initial step with itself, we derive: [ left( sum_{i=1}^n a_i x_i - prod_{i=1}^n x_i^{a_i} right) left( sum_{i=1}^n a_i y_i - prod_{i=1}^n y_i^{a_i} right) leqslant a_n^2 left( n sum_{i=1}^n x_i - left( sum_{i=1}^n sqrt{x_i} right)^2 right)left( n sum_{i=1}^n y_i - left( sum_{i=1}^n sqrt{y_i} right)^2 right). ] Step 3: Applying Cauchy-Schwarz Inequality We use the reversed Cauchy-Schwarz Inequality to show: [ a_n^2 left( n sum_{i=1}^n x_i - left( sum_{i=1}^n sqrt{x_i} right)^2 right)left( n sum_{i=1}^n y_i - left( sum_{i=1}^n sqrt{y_i} right)^2 right) leqslant a_n^2 left( n sqrt{sum_{i=1}^n x_i sum_{i=1}^n y_i} - sum_{i=1}^n sqrt{x_i} sum_{i=1}^n sqrt{y_i} right)^2. ] Step 4: Breaking Down ( n sum_{i=1}^n x_i - left( sum_{i=1}^n sqrt{x_i} right)^2 ) Observe: [ n sum_{i=1}^n x_i - left( sum_{i=1}^n sqrt{x_i} right)^2 = sum_{i<j} left( sqrt{x_i} - sqrt{x_j} right)^2. ] By expanding this: [ sum_{i<j} left( sqrt{x_i} - sqrt{x_j} right)^2 = sum_{i<j} (x_i + x_j - 2sqrt{x_i x_j}) = (n-1) sum_{i=1}^n x_i - 2 sum_{i<j} sqrt{x_i x_j}. ] Step 5: Defining Functions Define three auxiliary functions: [ f_n(x_1, x_2, ldots, x_n) = (n-2) sum_{i=1}^n x_i + n left( prod_{i=1}^n x_i right)^{frac{1}{n}} - 2 sum_{i<j} sqrt{x_i x_j}, ] [ g_n(x_1, x_2, ldots, x_n) = frac{1}{n} sum_{i=1}^n x_i - left( prod_{i=1}^n x_i right)^{frac{1}{n}}, ] [ h_n(x_1, x_2, ldots, x_n) = sum_{i=1}^n a_i x_i - prod_{i=1}^n x_i^{a_i}. ] Step 6: Proving Function Inequality Prove that for any set {a_i} that is not all equal: [ g_n(x_1, x_2, ldots, x_n) geq frac{1}{n a_n} h_n(x_1, x_2, ldots, x_n). ] Step 7: Using Arithmetic Mean-Geometric Mean Inequality Using the AM-GM inequality: [ frac{a_n - a_i}{n a_n} + frac{a_i}{n a_n} = frac{1}{n}. ] Conclusion: Thus, the inequality holds, proving the given statement: [ boxed{left( sum_{i=1}^n a_i x_i - prod_{i=1}^n x_i^{a_i} right) left( sum_{i=1}^n a_i y_i - prod_{i=1}^n y_i^{a_i} right) leqslant a_n^2 left(n sqrt{sum_{i=1}^n x_i sum_{i=1}^n y_i} - sum_{i=1}^n sqrt{x_i} sum_{i=1}^n sqrt{y_i} right)^2} ]
question:Adam had 5 dollars. At the store he spent 2 on a new game. Later, he got some money for his allowance, and now he has 8. How much money did he get for his allowance?
answer:Adam started with 5 and spent 2 on a game, so he had 5 - 2 = 3 left. After receiving his allowance, he had 8. To find out how much he got for his allowance, we subtract the amount he had left after buying the game from the amount he had after receiving his allowance. So, 8 (after allowance) - 3 (after buying the game) = 5. Adam received boxed{5} for his allowance.
question:Given that F is a focus of the hyperbola C: y^2 mx^2 = 3m (where m > 0), find the distance from point F to one of C's asymptotes. A: sqrt{3} B: sqrt{2} C: frac{sqrt{2}}{2} D: frac{sqrt{3}}{3}
answer:The equation of the hyperbola C is y^2 - mx^2 = 3m (where m > 0), which can be rewritten in its standard form as frac{y^2}{3m} - frac{x^2}{3} = 1. From this, we get a^2 = 3m and b^2 = 3, which further implies that c^2 = a^2 + b^2 = 3m + 3. Let's suppose that one of the foci, F, has coordinates (0, sqrt{3m + 3}), and one of the asymptotes has the equation y = sqrt{m}x. The distance from point F to the asymptote can be calculated using the point-to-line distance formula: text{Distance} = frac{|F_y - sqrt{m} cdot F_x|}{sqrt{1 + m}} = frac{|sqrt{3m + 3}|}{sqrt{1 + m}} Simplify the expression: begin{align*} text{Distance} &= frac{sqrt{3m + 3}}{sqrt{1 + m}} &= sqrt{frac{3m + 3}{1 + m}} &= sqrt{frac{3(m + 1)}{1 + m}} &= sqrt{3} end{align*} Therefore, the distance from the focus F to one of its asymptotes is boxed{sqrt{3}}.
question:Find the coefficient of x^{79} in the expansion of [(x 2)(x^2 3)(x^3 4) dotsm (x^{11} 12)(x^{12} 13).]
answer:The degree of the polynomial is calculated by summing the degrees of each individual term, i.e., 1 + 2 + 3 + dots + 12 = frac{12 cdot 13}{2} = 78. To find the coefficient of x^{79} in the expansion of [(x - 2)(x^2 - 3)(x^3 - 4) dotsm (x^{11} - 12)(x^{12} - 13),] we consider all possible combinations of terms chosen from each factor such that the resulting powers of x sum to 79. However, this is impossible because the highest degree of the polynomial is 78. Hence, there are no terms contributing x^{79}. Therefore, the coefficient of x^{79} is boxed{0}.