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Bernoulli's inequality can be proved for case 2, in which is a non-negative integer and , using mathematical induction in the following form: we prove the inequality for r ∈ { 0 , 1 } {\displaystyle r\in \{0,1\}} ,
Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes. [1] [2]Mathematical induction is a method for proving that a statement () is true for every natural number, that is, that the infinitely many cases (), (), (), (), … all hold.
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.
For the following proof we apply mathematical induction and only well-known rules of arithmetic. Induction basis: For n = 1 the statement is true with equality. Induction hypothesis: Suppose that the AM–GM statement holds for all choices of n non-negative real numbers. Induction step: Consider n + 1 non-negative real numbers x 1, . . . , x n+1, .
The proof is divided into three steps. The idea is to substitute the assumed integral inequality into itself n times. 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.
Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space L p (μ), and also to establish that L q (μ) is the dual space of L p (μ) for p ∈ [1, ∞). Hölder's inequality (in a slightly different form) was first found by Leonard James Rogers .
In mathematics, the rearrangement inequality [1] states that for every choice of real numbers ... For a proof by mathematical induction, we start with = ...
In proof by mathematical induction, a single "base case" is proved, and an "induction rule" is proved that establishes that any arbitrary case implies the next case. Since in principle the induction rule can be applied repeatedly (starting from the proved base case), it follows that all (usually infinitely many) cases are provable. [ 15 ]