Math IOQM

  Idea in plain words Line up two lists of numbers that both go up, like heights: a 1 ≤ a 2 ≤ ⋯ ≤ a n a_1 \le a_2 \le \dots \le a_n backpack weights: b 1 ≤ b 2 ≤ ⋯ ≤ b n b_1 \le b_2 \le \dots \le b_n Now pair the first height with the first weight, the second with the second, etc. Product of averages ( a 1 + ⋯ + a n n ) ( b 1 + ⋯ + b n n ) \displaystyle \left(\frac{a_1+\cdots+a_n}{n}\right)\left(\frac{b_1+\cdots+b_n}{n}\right) is like saying: “Pretend every student has the average height and carries the average backpack. What would the ‘typical’ height×weight be?” Average of products a 1 b 1 + a 2 b 2 + ⋯ + a n b n n \displaystyle \frac{a_1b_1+a_2b_2+\cdots+a_nb_n}{n} is what you actually get when the shorter students carry the lighter backpacks and the taller students carry the heavier ones. Chebyshev’s inequality says: when both lists rise together (small with small, big with big), average of products       ≥       product of a...

Power mean inequality: Power Mean  For real numbers x 1 , x 2 , … , x n ≥ 0 x_1, x_2, \ldots, x_n \ge 0 and real p ≠ 0 p \ne 0 , the power mean of order p p is: M p = ( x 1 p + x 2 p + ⋯ + x n p n ) 1 / p M_p = \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n} \right)^{1/p} Power Mean Inequality If p < q p < q , then: ( x 1 p + x 2 p + ⋯ + x n p n ) 1 / p ≤ ( x 1 q + x 2 q + ⋯ + x n q n ) 1 / q \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n} \right)^{1/p} \le \left( \frac{x_1^q + x_2^q + \cdots + x_n^q}{n} \right)^{1/q} with equality if and only if x 1 = x 2 = ⋯ = x n x_1 = x_2 = \cdots = x_n . Example (n = 3, p = 2, q = 3) Let’s say: x 1 , x 2 , x 3 ≥ 0 x_1, x_2, x_3 \ge 0 Then the power mean inequality says: ( x 1 2 + x 2 2 + x 3 2 3 ) 1 / 2 ≤ ( x 1 3 + x 2 3 + x 3 3 3 ) 1 / 3 \left( \frac{x_1^2 + x_2^2 + x_3^2}{3} \right)^{1/2} \le \left( \frac{x_1^3 + x_2^3 + x_3^3}{3} \right)^{1/3} Ex1) Prove that a 4 + b 4 + c 4 ≥ a b c ( a + b + c ) a^4 + b^4 + c^4...

  We want to prove: ∑ k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} Method1: Polynomial Assumption and Coefficient Matching The idea is to assume that the sum of squares is a cubic polynomial in n n : ∑ k = 1 n k 2 = A n 3 + B n 2 + C n + D \sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn + D But since ∑ k = 1 0 k 2 = 0 \sum_{k=1}^0 k^2 = 0 , we know that when n = 0 n = 0 , the sum is 0 ⇒ D = 0 D = 0 . So we assume: ∑ k = 1 n k 2 = A n 3 + B n 2 + C n \sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn Now compute the sum for small values of n n : n = 1 n = 1 : 1 2 = A ( 1 ) 3 + B ( 1 ) 2 + C ( 1 ) = A + B + C ⇒ A + B + C = 1 (1) 1^2 = A(1)^3 + B(1)^2 + C(1) = A + B + C \Rightarrow A + B + C = 1 \tag{1} n = 2 n = 2 : 1 2 + 2 2 = 1 + 4 = 5 ⇒ 8 A + 4 B + 2 C = 5 (2) 1^2 + 2^2 = 1 + 4 = 5 \Rightarrow 8A + 4B + 2C = 5 \tag{2} n = 3 n = 3 : 1 2 + 2 2 + 3 2 = 1 + 4 + 9 = 14 ⇒ 27 A + 9 B + 3 C = 14 (3) 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \Rightarrow 2...