关于不等式证明方法的探讨毕业论文(编辑修改稿)

关于不等式证明方法的探讨毕业论文(编辑修改稿)内容摘要：

x1 and x2. Similarly, 4√x1x2 is the perimeter of a square with the samearea. Thus for n = 2 the AM–GM inequality states that only the square has the smallest perimeter amongst all rectangles of equal area. The full inequality is an extension of this idea to n dimensions. Every vertex of an ndimensional box is connected to n edges. If these edges39。 lengths arex1, x2, . . . , xn, then x1 + x2 + + xn is the total length of edges incident to the vertex. There are 2n vertices, so we multiply this by 2n。 since each edge, however, meets two vertices, every edge is counted twice. Therefore we divide by 2 and conclude that there are 2n−1n edges. There are equally many edges of each length and n lengths。 hence there are 2n−1 edges of each length and the total edgelength is 2n−1(x1 + x2 + + xn). On the other hand, 1 122n n nn x x x is the total length of edges connected to a vertex on an ndimensional cube of equal volume. Since the inequality says 12 12 ,n n nx x x x x xn    we get 河北师范大学本科生毕业论文 111 2 1 22 ( ) 2nn nnnx x x n x x x    with equality if and only if x1 = x2 = = xn. Thus the AM–GM inequality states that only the ncube has the smallest sum of lengths of edges connected to each vertex amongst all ndimensional boxes with the same volume.[1] Example application Consider the function 3( , , ) x y zf x y z y z x   for all positive real numbers x, y and z. Suppose we wish to find the minimal value of this function. First we rewrite it a bit: 3331 2 3 4 5 61 1 1 1 12 2 3 3 3( , , ) 666 6x y y z z zy z z x x xf x y zx x x x x x         with 31 2 3 4 5 611, , .23x y zx x x x x xy z x      Applying the AM–GM inequality for n = 6, we get 333662 / 3 1 / 21 1 1 1 1( , , ) 62 2 3 3 3162 2 3 3 32 3 .x y y z z zf x y zy z z x x xx y zy z x          Further, we know that the two sides are equal exactly when all the terms of the mean are equal: 2 / 3 1 / 2 311( , , ) 2 3 w h e n .23x y zf x y z y z x    All the points (x, y, z) satisfying these conditions lie on a halfline starting at the origin and are given by 河北师范大学本科生毕业论文 3 33( , , ) , 2 3 , w i t h 0 .2()x y z x x x x Practical applications An important practical application in financial mathematics is to puting the rate of return: the annualized return, puted via the geometric mean, is less than the average annual return, puted by the arithmetic mean (or equal if all returns are equal). This is important in analyzing investments, as the average return overstates the cumulative effect. Proofs of the AM–GM inequality There are several ways to prove the AM–GM inequality。 for example, it can be inferred from Jensen39。 s inequality, using the concave function ln(x). It can also be proven using the rearrangement inequality. Considering length and required prerequisites, the elementary proof by induction given below is probably the best remendation for first reading. Idea of the first two proofs We have to show that 12 12n n nx x x x x xn    with equality only when all numbers are equal. If xi ≠ xj, then replacing both xi and xj by (xi + xj)/2 will leave the arithmetic mean on the lefthand side unchanged, but will increase the geometric mean on the righthand side because 22 ( ) ( )i j i jijx x x xxx   Thus righthand side will be largest — so the idea — when all xis are equal to the arithmetic mean 12 ,nx x xn   thus as this is then the largest value of righthand side of the expression, we have 12 12 .nn n nx x x x x xn           This is a valid proof for the case n = 2, but the procedure of taking iteratively pairwise averages may fail to produce n equal numbers in the case n ≥ 3. An example of this case is x1 = x2 ≠ x3: Averaging two different numbers produces two equal numbers, but the third one is still different. Therefore, we never actually get an inequality involving the geometric mean of three equal numbers. Hence, an additional trick or a modified argument is necessary to turn the above idea into a valid proof for the case n ≥ 3. 河北师范大学本科生毕业论文 Proof by induction With the arithmetic mean 1 nxxn  of the nonnegative real numbers x1, . . . , xn, the AM–GM statement is equivalent to 12n nx x x  with equality if and only if α = xi for all i ∈ {1, . . . , n}. For the following proof we apply mathematical induction and only wellknown 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 nonnegative real numbers. Induction step: Consider n +1 nonnegative real numbers. Their arithmetic mean α satisfies. 河北师范大学本科生毕业论文 本科生毕业论文设计 关于不等式证明方法的探讨 作 者 姓 名 ： 曾海辉 指 导 教 师 ： 张硕 所 在 学 院 ： 数学与信息科学学院 专 业 （ 系 ）： 数学与应用数学专业 班 级 （ 届 ）： 2020 届数学 B 班 二〇一四年 四月 三十日河北师范大学本科生毕业论文 1 目录 目录 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„1 摘要、关键字 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 2 1 问题提出 „„„„„ „„„„„„„„„„„„„„„„„„„„„„„„„„„ 3 在现实生活中的意义及前景 „„„„„„„„„„„„„„„„„„„„„„ 3 在数学教学中的现状和问题 „„„„„„„„„„„„„„„„„„„„„„ 3 2 常 用证明方法 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 5 比较法 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 5 分析综合法 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 6 反证法 „„„„ „„„„„„„„„„„„„„„„„„„„„„„„„„„ 7 放缩法 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 8 换元法 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 11 数学归纳法 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 14 判别式法 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 15 函数单调性法 „„„„„„„„„„„„„„„„„„„„„„„„„„„„ 16 几何证法 „ „„„„„„„„„„„„„„„„„„„„„„„„ „„„„ „ 17 面积体积法 „ „„„ „„„„„„„„„„„„„„„„„„„„„„„„ 18 极值法 „„„„„„„„„„„„„„„„„„„„„„„„„„ „„„ „ 19 3 教学建议与思考 „„„„„„„„„„„„„„„„„„„„„„„„„ „„ „„ 20 内容综述与建议 „„„„„„„„„„„„„„„„„„„„„„„„„„„ 1。