# Quod Erat Demonstrandum

## 2008/12/03

### 證明不等式的基礎招式 (Part 2)

Filed under: Pure Mathematics,Teaching — johnmayhk @ 5:12 下午
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e.g. 11 For any positive integer $n$, show that $(1 + \frac{1}{n + 1})^{n + 1} > (1 + \frac{1}{n})^n$

$1$$1$$n$$1 + \frac{1}{n}$

$\frac{1 + n(1 + \frac{1}{n})}{n + 1} > \sqrt[n + 1]{1 \times (1 + \frac{1}{n})^n}$
$\Rightarrow \frac{1 + n + 1}{1 + n} > \sqrt[n + 1]{(1 + \frac{1}{n})^n}$
$\Rightarrow (1 + \frac{1}{n + 1})^{n + 1} > (1 + \frac{1}{n})^n$

e.g. 12 For any positive integers $m, n, p$ and non-negative numbers $a, b, c$, show that $(\frac{ma + nb + pc}{m + n + p})^{m + n + p} \ge a^mb^nc^p$.

e.g. 13
(a) $1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$

(b) $1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$

(c) $1^3 + 2^3 + 3^3 + \dots + n^3 = \frac{n^2(n + 1)^2}{4}$

(d) $1 + a + a^2 + \dots + a^{n - 1} = \frac{a^n - 1}{a - 1}$ ($a \ne 1$)
$\Rightarrow \frac{a^n - 1}{n(a - 1)} > \sqrt[n]{a^{1 + 2 + 3 + \dots + (n - 1)}}$

(e) $1 + 2a + 3a^2 + \dots + na^{n - 1} = \frac{1 - (n + 1)a^n + na^{n + 1}}{(1 - a)^2}$ ($a \ne 1$)

(f) $C_0^n + C_1^n + C_2^n + \dots + C_n^n = 2^n$

$(C_1^n)^2 + 2(C_2^n)^2 + 3(C_3^n)^2 + \dots + n(C_n^n)^2 = \frac{(2n - 1)!}{[(n - 1)!]^2}$

(g) $\frac{1}{1\times 2} + \frac{1}{2\times 3} + \dots + \frac{1}{n\times (n + 1)} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n + 1}) = 1 - \frac{1}{n + 1}$
$\Rightarrow \frac{1}{n}(1 - \frac{1}{n + 1}) > \sqrt[n]{\frac{1}{1 \times 2}\frac{1}{2 \times 3}\dots\frac{1}{n(n + 1)}}$
$\Rightarrow \frac{1}{n + 1} > \sqrt[n]{\frac{1}{(n!)^2(n + 1)}}$

$(a_1^2 + a_2^2 + \dots + a_n^2)(b_1^2 + b_2^2 + \dots + b_n^2) \ge (a_1b_1 + a_2b_2 + \dots + a_nb_n)^2$

e.g. 14 For non-negative numbers $x_1, x_2, \dots ,x_n$, show that $n(x_1 + x_2 + \dots + x_n) \ge (\sqrt{x_1} + \sqrt{x_2} + \dots + \sqrt{x_n})^2$

$(1 + 1 + \dots + 1)(x_1 + x_2 + \dots + x_n)$
$= (1^2 + 1^2 + \dots + 1^2)(\sqrt{x_1}^2 + \sqrt{x_2}^2 + \dots + \sqrt{x_n}^2)$
$\ge (1\times\sqrt{x_1} + 1\times\sqrt{x_2} + \dots + 1\times\sqrt{x_n})^2$

result follows.

$n$ 是平方數，比方設 $n = 9$，我們可進一步得到：$\sqrt{x_1 + x_2 + \dots + x_9} \ge \frac{\sqrt{x_1} + \sqrt{x_2} + \dots + \sqrt{x_9}}{3}$

e.g. 15 Show that $r.m.s. \ge A.M.$, that is $\sqrt{\frac{x_1^2 + x_2^2 + \dots + x_n^2}{n}} \ge \frac{x_1 + x_2 + \dots + x_n}{n}$

$(1^2 + 1^2 + \dots + 1^2)(x_1^2 + x_2^2 + \dots x_n^2) \ge (x_1 + x_2 + \dots + x_n)^2$

e.g. 16 Show that $\sqrt{8^n - 4^n - 2^n + 1} \ge \sqrt{3}\sum_{k = 1}^n\sqrt{4^{n - k}C_k^n}$

$_nC_1 + _nC_2 + \dots + _nC_n = 2^n - 1$
$1 + a^2 + a^4 + \dots + a^{2n - 2} = \frac{a^{2n} - 1}{a^2 - 1}$ ($a \ne 1$)

$(_nC_1 + _nC_2 + \dots + _nC_n)(1 + a^2 + a^4 + \dots + a^{2n - 2})$
$\ge (\sqrt{_nC_1} + a\sqrt{_nC_2} + a^2\sqrt{_nC_3} + \dots a^{n - 1}\sqrt{_nC_n})^2$

$\Rightarrow (2^n - 1)(\frac{a^{2n} - 1}{a^2 - 1}) \ge (\sqrt{_nC_1} + a\sqrt{_nC_2} + a^2\sqrt{_nC_3} + \dots a^{n - 1}\sqrt{_nC_n})^2$

$(2^n - 1)(\frac{4^n - 1}{3}) \ge (\sum_{k = 1}^n2^{k - 1}\sqrt{_nC_k})^2$
$\Rightarrow (2^n - 1)(4^n - 1) \ge 3(\sum_{k = 1}^n\sqrt{4^{k - 1}C_k^n})^2$
$\Rightarrow \sqrt{8^n - 4^n - 2^n + 1} \ge \sqrt{3}\sum_{k = 1}^n\sqrt{4^{k - 1}C_k^n}$

$\Rightarrow \sqrt{8^n - 4^n - 2^n + 1} \ge \sqrt{3}\sum_{k = 1}^n\sqrt{4^{n - k}C_k^n}$

e.g. 17 Given that $\frac{a + b + c + d}{4} \ge \sqrt[4]{abcd}$ for any non-negative numbers $a, b, c, d$, show that $\frac{a + b + c}{3} \ge \sqrt[3]{abc}$.

$\frac{3d + d}{4} \ge \sqrt[4]{abcd}$
$\Rightarrow d^4 \ge abcd$
$\Rightarrow d^3 \ge abc$
$\Rightarrow d \ge \sqrt[3]{abc}$
$\Rightarrow \frac{a + b + c}{3} \ge \sqrt[3]{abc}$

e.g. 18 Given that for any positive numbers $a_1, a_2, \dots ,a_n$ such that $\sum_{k = 1}^n a_k = 1$, then $\sum_{k = 1}^{n} a_k^p \le n\sum_{k = 1}^{n} a_k^{p + 1}$ for any non-negative integer $p$. Show that, for any positive numbers $b_1, b_2, \dots ,b_n$, $(\sum_{k = 1}^n b_k)(\sum_{k = 1}^n b_k^p) \le n\sum_{k = 1}^n b_k^{p + 1}$ for any non-negative integer $p$.

$\sum_{k = 1}^n b_k = B$ (say)

$\sum_{k = 1}^n \frac{b_k}{B} = 1$

$\sum_{k = 1}^{n} (\frac{b_k}{B})^p \le n\sum_{k = 1}^{n} (\frac{b_k}{B})^{p + 1}$
$\Rightarrow \frac{\sum_{k = 1}^{n} b_k^p}{B^p} \le n\frac{\sum_{k = 1}^{n} b_k{p + 1}}{B^{p + 1}}$
$\Rightarrow B\sum_{k = 1}^{n} b_k^p \le n\sum_{k = 1}^{n} b_k^{p + 1}$
$\Rightarrow (\sum_{k = 1}^n b_k)(\sum_{k = 1}^n b_k^p) \le n\sum_{k = 1}^n b_k^{p + 1}$ for any non-negative integer $p$

e.g. 19 Given that for any positive numbers $a_1, a_2, \dots ,a_n$ such that $a_1a_2 \dots a_n = 1$, then $a_1 + a_2 + \dots + a_n \ge n$. Show that, for any positive numbers $b_1, b_2, \dots ,b_n$, we have $\frac{b_1 + b_2 + \dots + b_n}{n} \ge \sqrt[n]{b_1b_2 \dots b_n}$

$b_1b_2 \dots b_n = B$ (say)

$\frac{b_1b_2 \dots b_n}{B} = 1$
$\frac{b_1}{\sqrt[n]{B}}\frac{b_2}{\sqrt[n]{B}}\dots\frac{b_n}{\sqrt[n]{B}} = 1$ [即把 $B$ 平均地分配給 $b_1, b_2, \dots ,b_n$]

$\frac{b_1}{\sqrt[n]{B}} + \frac{b_2}{\sqrt[n]{B}} + \dots + \frac{b_n}{\sqrt[n]{B}} = n$
$\Rightarrow \frac{b_1 + b_2 + \dots b_n}{n} \ge \sqrt[n]{B}$
$\Rightarrow \frac{b_1 + b_2 + \dots b_n}{n} \ge \sqrt[n]{b_1b_2 \dots b_n}$

e.g. 20 Given that for any positive numbers $a_1, a_2, \dots ,a_n$ and $b_1, b_2, \dots ,b_n$ such that $\sum_{k = 1}^n a_kb_k = \sum_{k = 1}^n b_k = 1$, then $\prod_{k = 1}^na_k^{b_k} \le 1$.

Show that for any positive numbers $c_k, d_k$ ($k = 1, 2, \dots ,n$),

$(\prod_{k = 1}^nc_k^{d_k})^{\frac{1}{\sum_{k = 1}d_k}} \le \frac{\sum_{k = 1}^n c_kd_k}{\sum_{k = 1}^n d_k}$.

$\prod [\frac{c_k\sum d_k}{\sum_{c_kd_k}}]^{\frac{d_k}{\sum d_k}} \le 1$

（為簡化起見，我不寫上下標。）

$\prod c_k^{\frac{d_k}{\sum d_k}} \times \prod [\frac{\sum d_k}{\sum c_kd_k}]^{\frac{d_k}{\sum d_k}} \le 1$

$(\prod c_k^{d_k})^{\frac{1}{\sum d_k}} \times [\frac{\sum d_k}{\sum c_kd_k}]^{\frac{\sum d_k}{\sum d_k}} \le 1$

result follows.

## 1 則迴響 »

1. Hi,

It will be nice to have an English equivalent of your two articles to reach out to more people :)

Thanks!

Cheers,
Wen Shih

迴響 由 Wen Shih — 2008/12/08 @ 10:29 下午 | 回覆