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Prove the uniqueness of the given solution: this step implies the physical correctness of the problem, showing that the solution can be used to represent a physical phenomenon. It is a particularly important step since most of the problems modeled by variational inequalities are of physical origin. Find the solution or prove its regularity.
This is done in Claim 1 using mathematical induction. In Claim 2 we rewrite the measure of a simplex in a convenient form, using the permutation invariance of product measures. In the third step we pass to the limit n to infinity to derive the desired variant of Grönwall's inequality.
Computer support in solving inequations is described in constraint programming; in particular, the simplex algorithm finds optimal solutions of linear inequations. [6] The programming language Prolog III also supports solving algorithms for particular classes of inequalities (and other relations) as a basic language feature.
There are three inequalities between means to prove. There are various methods to prove the inequalities, including mathematical induction, the Cauchy–Schwarz inequality, Lagrange multipliers, and Jensen's inequality. For several proofs that GM ≤ AM, see Inequality of arithmetic and geometric means.
[2] Whether a space supports a Poincaré inequality has turned out to have deep connections to the geometry and analysis of the space. For example, Cheeger has shown that a doubling space satisfying a Poincaré inequality admits a notion of differentiation. [3] Such spaces include sub-Riemannian manifolds and Laakso spaces.
Since all the inequalities are in the same form (all less-than or all greater-than), we can examine the coefficient signs for each variable. Eliminating x would yield 2*2 = 4 inequalities on the remaining variables, and so would eliminating y. Eliminating z would yield only 3*1 = 3 inequalities so we use that instead.
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For instance, to solve the inequality 4x < 2x + 1 ≤ 3x + 2, it is not possible to isolate x in any one part of the inequality through addition or subtraction. Instead, the inequalities must be solved independently, yielding x < 1 / 2 and x ≥ −1 respectively, which can be combined into the final solution −1 ≤ x < 1 / 2 .