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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.
Throughout the proof, given and , we shall assume that =. A double induction argument is applied to the couple of integers (,), representing the orders of differentiation. The other parameters are constructed in such a way that they comply with the hypotheses of the theorem.
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.
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\}} ,
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 ]
The proof of the general Leibniz rule [2]: 68–69 proceeds by induction. Let f {\displaystyle f} and g {\displaystyle g} be n {\displaystyle n} -times differentiable functions. The base case when n = 1 {\displaystyle n=1} claims that: ( f g ) ′ = f ′ g + f g ′ , {\displaystyle (fg)'=f'g+fg',} which is the usual product rule and is known ...
Bregman–Minc inequality (discrete mathematics) Brianchon's theorem ; British flag theorem (Euclidean geometry) Brooks's theorem (graph theory) Brouwer fixed-point theorem ; Browder–Minty theorem (operator theory) Brown's representability theorem (homotopy theory) Bruck–Chowla–Ryser theorem (combinatorics) Brun's theorem (number theory)
The inequality with the subtractions can be proven easily via mathematical induction. The one with the additions is proven identically. The one with the additions is proven identically. We can choose n = 1 {\displaystyle n=1} as the base case and see that for this value of n {\displaystyle n} we get