Schur inequality shows that for all non-negative real numbers x, y, z and positive number t, there are: if x, y and z & gt=0, then ∑ (x t) (x-y) > = 0 equals "=" if and only if x = y = z, or if two numbers are equal and the other is zero. When t is a positive even number, the inequality proves the Schur inequality for all real numbers x, y and z: suppose x> = y> = z ∑ x (x-y) (x-z) = x (x-y)+y (y-x)+z (y-z). = x (x-y) (x-z)+y (y-x) (y-z) > = x (x-y) (y-z)+y (y-x) = (x-y) 2 (y-z) > In fact, when t is an arbitrary real number, we can still prove Schur inequality. Schur inequality is not an inequality stipulated in the league syllabus, but it can still play an important role in the proof of league inequality. Held inequality is an inequality in mathematical analysis, which is based on Otto H? Lder). This is a basic inequality that reveals the relationship between L p spaces: let S be a measure space, let F be in L p (S), and let G be in L q (S). Then f g is within L 1 (S), and there is. If s is {1, ..., n}, a special case of Holder inequality is obtained by counting measure: for all real numbers (or complex n；umbers) x 1, ..., x n; Y 1, ... Yes, yes. We call P and Q the shackles of the holders. If we take S as the counting measure of natural number set, we will get an infinite series inequality similar to that on. When p = q = 2, Cauchy-Schwartz inequality is obtained. Holder inequality can prove the generalized triangle inequality and Minkowski inequality in L p space, and prove that L p space is the duality of L q space. [Editor] Note: In the definition of held yoke, 1/∞ means zero. If 1 ≤ p, q is multiplied by ∞, and ∞ is obtained. [Editor] There are many proofs of Holder's inequality, and the main idea is Young's inequality. If ||f || p = 0, f μ- is almost everywhere and the product fg μ- is almost everywhere, so the left end of Holder inequality is zero. If || g ||g|| q = 0, the same holds true. Therefore, we can assume that ||f || p > 0 and ||| g || q > 0. If ||||| f ||| p =∞ or ||| g ||||| q =∞, then the right end of the inequality is infinite. Therefore, we can assume that ||f || p and || g ||g|| q lie within (0, ∞). If p = ∞ and q = 1, then |fg| ≤ ||f || ∞ |g| exists almost everywhere, and the inequality can be deduced from the monotonicity of Lebesgue integral. For p = 1 and q = ∞, the situation is similar. So we can also assume p, q ∈ (1, ∞). Divide ||||| f |||||||| by f and g respectively, and we can assume that we are using Young's inequality now: For all nonnegative A and B, the equation holds if and only if A P = B Q. Therefore: two-sided integration, so: This proves holder inequality. Under the assumptions of p ∈ (1, ∞) and |||||||| p =|||||| q =1,the equation exists almost everywhere if and only if | f | p =| g | q. More generally, if ||| f ||| p and ||| g ||g|| q lie within (0, ∞), then the held inequality becomes an equation if and only if α, β >; 0 (that is, α = ||||| g ||||| q and β = ||||| p), so: μ-almost everywhere (*) ||| f |||| p = 0 corresponds to β = 0 in (*). ||| g ||g|| q = α = 0 in the corresponding (*). [Editor] Refer to Hardy and G.H.; Littlewood & ampG. Pólya (1934), Inequalities, Cambridge University Press, ISBN 052 1358809 H? Lder, O. (1889), "Ueber einen Mittelwerthsatz", National Human Rights Commission. Ges。 Weiss. g？ Tingen: 38–47 Kuptsov, L.P. (200 1), "H? Lder inequality ",in Hazewinkel, Michiel, Encyclopedia of Mathematics, Cromwell Academic Press, ISBN 978-1556080104 Rogers, L. j .( 1888)," Extension of a theorem in inequality ", Mathematical Messenger17:145–150 Kuttler, Kenneth (2007), Introduction to Linear Algebra, online electronic version visited on June 27th, 2009. Zhang Yuanzhang, Proof and Application of Young's Inequality, Henan Science. No.0 1, vol. 22. Visited on June 27th, 2009. Excerpted from "/%E8% B5% AB"