Table of Contents

  1. Polya’s Coaching and Carleman’s Inequality
  2. exercise
    1. Bounds by Pure Powers
    2. A Canadian Challenge
    3. A Bound Between Differences
    4. Akerberg’s Refinement
    5. An AM-GM Inequality for Complex Numbers
    6. A Leap-Forward Fall-Back Tour de Force

对非负实数 $ a_ {1}, a_ {2}, \ldots, a_ {n} $ 和每个正实数序列 $ p_ {1}, p_ {2}, \ldots, p_ {n} $,该序列和为1,有

$ \begin{equation} a_ {1}^{p_ {1}} a_ {2}^{p_ {2}} \ldots a_ {n}^{p_ {n}} \leq p_ {1}a_ {1} + p_ {2}a_ {2} + \cdots + p_ {n}a_ {n} \end{equation} $

证明:我们利用公式 $ x \leq e^{x-1} \qquad for \, all \, x \in R $,有

$ \begin{equation} a_ {k} \leq e^{a_ {k} - 1} \quad and \quad a_ {k}^{p_ {k}} \leq e^{p_ {k}a_ {k} - p_ {k}} \end{equation} $

则有 $ a_ {1}^{p_ {1}} a_ {2}^{p_ {2}} \ldots a_ {n}^{p_ {n}} = exp\left(\{\sum_ {k=1}^{n}p_ {k}a_ {k} \} - 1 \right) $

然后我们考虑引入新变量 $ \alpha_ {k}, k = 1,2,\ldots, n $,使得

$ \begin{equation} \alpha_ {k} = \frac{a_ {k}}{A} \qquad where \, A = p_ {1}a_ {1} + p_ {2}a_ {2} + \cdots + p_ {n}a_ {n} \end{equation} $

然后我们得到 $ \left(\frac{a_ {1}}{A}\right)^{p_ {1}} \left(\frac{a_ {2}}{A}\right)^{p_ {2}} \cdots \left(\frac{a_ {n}}{A}\right)^{p_ {n}} \leq exp\left(\{\sum_ {k=1}^{n}p_ {k} \frac{a_ {k}}{A} \} - 1\right) = 1 $

则得证。

Polya’s Coaching and Carleman’s Inequality

Carleman’s Inequality 对每个正实数序列 $ a_ {1}, a_ {2}, \ldots $,有

$ \begin{equation} \sum_ {k=1}^{\infty} \left( a_ {1} a_ {2} \cdots a_ {k} \right)^{\frac{1}{k}} \leq e \sum_ {k=1}^{\infty}a_ {k} \end{equation} $

证明:这里我们构造一序列因子ck ,有

$ \begin{equation} \begin{aligned} \sum_ {k=1}^{\infty}\left(a_ {1} a_ {2} \cdots a_ {k}\right)^{\frac{1}{k}} &= \sum_ {k=1}^{\infty}\frac{\left(a_ {1}c_ {1}a_ {2}c_ {2} \cdots a_ {k}c_ {k}\right)^{\frac{1}{k}}}{\left(c_ {1}c_ {2} \cdots c_ {k}\right)^{\frac{1}{k}}} \\ &\leq \sum_ {k=1}^{\infty} \sum_ {k=1}{\infty}\frac{a_ {1}c_ {1} + a_ {2}c_ {2} + \cdots + a_ {k}c_ {k}}{k\left(c_ {1}c_ {2} \cdots c_ {k}\right)^{\frac{1}{k}}} \\ &=\sum_ {k=1}^{\infty}a_ {k}c_ {k}\sum_ {j=k}^{\infty}\frac{1}{j\left(c_ {1}c_ {2} \cdots c_ {j}\right)^{\frac{1}{j}}} \end{aligned} \end{equation} $

我们可以精心选择这样的ck ,$ k = 1, 2, \ldots $,使得其和有界

$ \begin{equation} s_ {k} = c_ {k} \sum_ {j=k}^{\infty}\frac{1}{j\left(c_ {1} c_ {2} \cdots c_ {j}\right)^{\frac{1}{j}}} \qquad k = 1, 2, \ldots \end{equation} $

于是我们取这样的ck ,使得

$ \begin{equation} \left( c_ {1} c_ {2} \cdots c_ {j}\right)^{\frac{1}{j}} = j + 1 \qquad for \, j = 1, 2, \ldots \end{equation} $

这样我们有

$ \begin{equation} s_ {k} = c_ {k} \sum_ {j=k}^{\infty}\frac{1}{j \left(c_ {1}c_ {2} \cdots c_ {j}\right)^{\frac{1}{j}}} = c_ {k} \sum_ {j=k}^{\infty}\frac{1}{j\left(j + 1\right)} = \frac{c_ {k}}{k} \end{equation} $

通过如下等式:

$ \begin{equation} c_ {1} c_ {2} \cdots c_ {j-1} = j^{j-1} \quad and \quad c_ {1} c_ {2} \cdots c_ {j} = \left(j+1\right)^{j} \end{equation} $

我们可以推出:

$ \begin{equation} c_ {j} = \frac{\left(j+1\right)^{j}}{j^{j-1}} = j \left(1 + \frac{1}{j}\right)^{j} \end{equation} $

这样我们就得到:

$ \begin{equation} \sum_ {k=1}^{\infty}\left(a_ {1} a_ {2} \cdots a_ {k}\right)^{\frac{1}{k}} \leq \sum_ {k=1}^{\infty}\left(1 + \frac{1}{k}\right)^{k} a_ {k} \leq e \sum_ {k=1}^{\infty}a_ {k} \end{equation} $

exercise

Bounds by Pure Powers

试证明对正实数x, y, $ \alpha, \beta $,有

$ \begin{equation} x^{\alpha}y^{\beta} \leq \frac{\alpha}{\alpha+\beta}x^{\alpha+\beta} + \frac{\beta}{\alpha+\beta}y^{\alpha+\beta} \end{equation} $

对典型地推理,证明更好的边界:$ x^{2004}y + xy^{2004} \leq x^{2005}+y^{2005} $

证明:利用公式 $ a_ {1}^{p_ {1}}a_ {2}^{p_ {2}} \leq p_ {1}a_ {1} + p_ {2}a_ {2} $,我们设 $p_ {1} = \frac{\alpha}{\alpha+\beta}, p_ {2} = \frac{\beta}{\alpha+\beta}, a_ {1} = x^{\alpha+\beta}, a_ {2} = y^{\alpha + \beta} $,然后利用该不等式即得证。

然后利用求得的公式,令 $ \alpha = 2004, \beta = 1 \, and \, \alpha = 1, \beta = 2004 $带入分别计算,然后加和可得后面不等式的证明。

A Canadian Challenge

2002年加拿大数学奥林匹克赛中参与者被要求证明如下公式边界

$ \begin{equation} a + b + c \leq \frac{a^{3}}{bc} + \frac{b^{3}}{ac} + \frac{c^{3}}{ab} \end{equation} $

并确定等号什么时候成立。

证明:先把不等式两边乘以abc,得到 $ a^{2}bc + ab^{2}c + abc^{2} \leq a^{4} + b^{4} + c^{4} $,再同样利用之前学过的不等式,我们有 $ a^{2}bc = \left(a^{4}\right)^{\frac{1}{2}} + \left(b^{4}\right)^{\frac{1}{4}} + \left(c^{4}\right)^{\frac{1}{4}} \leq \frac{1}{2} a^{4} + \frac{1}{4}b^{4} + \frac{1}{4}c^{4} $,$ ab^{2}c, abc^{2} $也同样处理,最后相加即得证

等式成立需要 $ a = b = c $

A Bound Between Differences

对非负实数x和y及整数n有 $ n \left(x - y\right)\left(xy\right)^{\frac{\left(n-1\right)}{2}} \leq x^{n} - y^{n} $

证明:利用不等式 $ \left(x^{j+k}y^{j+k}\right)^{\frac{1}{2}} \leq \frac{1}{2} \left(x^{j}y^{k}+x^{k}y^{j}\right) $,我们设 k = n - 1 - j且对 $ 0 \leq j < n $遍历有:

$ \begin{equation} n\left(x+y\right)^{\frac{\left(n-1\right)}{2}} \leq x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1} = \frac{x^{n} - y^{n}}{x-y} \end{equation} $

即得证

Akerberg’s Refinement

对任意非负实数 $ a_ {1}, a_ {2}, \ldots, a_ {n} $ 且 $ n \ge 2 $,有如下边界:

$ \begin{equation} a_ {n}\{\frac{a_ {1} + a_ {2} + \cdots + a_ {n-1}}{n-1}\}^{n-1} \le \{ \frac{a_ {1} + a_ {2} + \cdots + a_ {n}}{n}\}^{n} \end{equation} $

该关系是AM-GM不等式的提炼因为AM-GM不等式符合该边界的迭代。为证明该式,首先必须证明

$ \begin{equation} y\left(n - y^{n-1}\right) = ny - y^{n} \le n - 1 \qquad \forall y \ge 0 \end{equation} $

其中的关键是选择一个巧妙的y

证明:第二个不等式很容易证明,这里不详细展开了。设 $ S_ {n} = a_ {1} + a_ {2} + \cdots + a_ {n} $,然后我们选择y使得 $ y^{n-1} = \frac{n a_ {n}}{S_ {n}} $,则我们有 $ n - y^{n-1} = \frac{n S_ {n-1}}{S_ {n}} $,然后带入第二个不等式,有 $ \left(\frac{n a_ {n}}{S_ {n}}\right)^{\frac{1}{n-1}} \frac{n S_ {n-1}}{S_ {n}} \le n - 1 $,调整相关项得 $ \left(\frac{n a_ {n}}{S_ {n}}\right)^{\frac{1}{n-1}} \frac{S_ {n-1}}{n-1} \le \frac{S_ {n}}{n} $,然后两边同取n - 1次方得 $ \frac{n a_ {n}}{S_ {n}} \left(\frac{S_ {n-1}}{n-1}\right)^{n-1} \leq \left(\frac{S_ {n}}{n}\right)^{n-1} $,然后整理一下即得证

An AM-GM Inequality for Complex Numbers

考虑n个复数 $ z_ {1}, z_ {2}, \ldots, z_ {3} $的集合S,其极坐标形式 $ z_ {j} = \rho_ {j}e^{i \theta_ {j}} $满足如下限制

$ \begin{equation} 0 \le \rho_ {j} < \infty \qquad and \qquad 0 \le | \theta_ {j} | < \psi < \frac{\pi}{2} \end{equation} $

img

如图所示,参数 $ z_ {j} \in S $的范围在 $ 2 \psi $,证明对这样的数有一下边界:

$ \begin{equation} \left(cos \psi\right) | z_ {1} z_ {2} \cdots z_ {n} |^{\frac{1}{n}} \le \frac{1}{n} | z_ {1} + z_ {2} + \cdots + z_ {n} | \end{equation} $

注意如果 $ z_ {j}, j = 1,2,\ldots, n $都是实数的话,则 $ \psi = 0 $,该边界即为AM-GM不等式。

证明:因 $ | w | \ge | \Re(w) | $且 $ \Re(z_ {j}) = \rho_ {j} \cos(\theta_ {j}) $,我们有

$ \begin{equation} \begin{aligned} | z_ {1} + z_ {2} + \cdots + z_ {n} | &\ge | \Re(z_ {1} + z_ {2} + \cdots + z_ {n}) | \\ &= | z_ {1} | \cos(\theta_ {1}) + | z_ {2} |\cos(\theta_ {2}) + \cdots + | z_ {n} |\cos(\theta_ {n}) \\ &\ge \left(| z_ {1} | + | z_ {2} | + \cdots + | z_ {n} |\right)\cos(\psi) \\ &\ge n\left(|z_ {1} | | z_ {2} | \cdots | z_ {n} |\right)^{\frac{1}{n}} \cos(\psi) \end{aligned} \end{equation} $

A Leap-Forward Fall-Back Tour de Force

使用Cauchy的前倾后退方法推导证明对任意非负数 $ x_ {1}, x_ {2}, \ldots, x_ {n} $及所有整数次方 $ n = 1,2,\ldots $有如下边界:

$ \begin{equation} \{ \frac{x_ {1} + x_ {2} + \cdots + x_ {m}}{m} \}^{n} \le \frac{x_ {1}^{n} + x_ {2}^{n} + \cdots + x_ {m}^{n}}{m} \end{equation} $

证明:考虑公式 $ H\left(n\right) : \left(\frac{x+y}{2}\right)^{n} \le \frac{x^{n} + y^{n}}{2}, \forall x \ge 0, y \ge 0 $,我们来推导H(n+1):

$ \begin{equation} \begin{aligned} \left(\frac{x+y}{2}\right)^{n+1} &= \left(\frac{x+y}{2}\right) \left(\frac{x+y}{2}\right)^{n} \le \left(\frac{x+y}{2}\right) \frac{x^{n} + y^{n}}{2} \\ &= \frac{x^{n+1} + y^{n+1} + xy^{n} + yx^{n}}{4} \\ &= \frac{x^{n+1} + y^{n+1}}{2} - \frac{\left(x - y\right)\left(x^{n} - y^{n}\right)}{4} \le \frac{x^{n+1} + y^{n+1}}{2} \end{aligned} \end{equation} $

这样对所有 $ n \ge 1 $ H(n)都有效。

现在我们这样处理H(n):

$ \begin{equation} \begin{aligned} \{\frac{x_ {1} + x_ {2} + x_ {3} + x_ {4}}{4} \}^{n} &\le \frac{1}{2} \{ \left(\frac{x_ {1} + x_ {2}}{2}\right)^{n} + \left(\frac{x_ {3} + x_ {4}}{2}\right)^{n} \} \\ &\le \frac{1}{2} \{ \frac{x_ {1}^{n} + x_ {2}^{n}}{2} + \frac{x_ {3}^{n} + x_ {4}^{n}}{2} \} \\ &= \frac{x_ {1}^{n} + x_ {2}^{n} + x_ {3}^{n} + x_ {4}^{n}}{4} \end{aligned} \end{equation} $

这样重复参数使得对每个k和 $ 2^{k} $个非负实数 $ x_ {1}, x_ {2}, \ldots, x_ {2^{k}} $的集合,我们有

$ \begin{equation} \{ \frac{x_ {1} + x_ {2} + \cdots + x_ {2^{k}}}{2^{k}} \}^{n} \le \frac{x_ {1}^{n} + x_ {2}^{n} + \cdots + x_ {2^{k}}^{n}}{2^{k}} \end{equation} $

这里我们设 $ H_ {new}\left(m\right) $为如下公式:

$ \begin{equation} \{ \frac{x_ {1} + x_ {2} + \cdots + x_ {m}}{m} \}^{n} \le \frac{x_ {1}^{n} + x_ {2}^{n} + \cdots + x_ {m}^{n}}{m} \end{equation} $

m为非负实数 $ x_ {1}, x_ {2}, \ldots, x_ {m} $的任意集合,我们知道 $ H_ {new}\left(m\right) $当m为2的次方是有效,这里我们如果能证明当$ H_ {new}\left(m\right) $ 有效时$ H_ {new}\left(m - 1\right) $也有效即可。

给定m - 1个非负实数 $ S = \{ x_ {1}, x_ {2}, \ldots, x_ {m-1} $,我们引入一个新的变量y,$ y = \frac{x_ {1} + x_ {2} + \cdots + x_ {m-1}}{m-1} $,因为y和 $ \frac{x_ {1} + x_ {2} + \cdots + x_ {m-1} + y}{m} $相等,我们应用H(m)到集合 $ S \cup \{y\} $,有

$ \begin{equation} y^{n} \le \frac{x_ {1}^{n} + x_ {2}^{n} + \cdots + x_ {m-1}^{n} + y^{n}}{m} \end{equation} $

整理可得

$ \begin{equation} y^{n} \le \frac{x_ {1}^{n} + x_ {2}^{n} + \cdots + x_ {m-1}^{n}}{m-1} \end{equation} $

即得证