# Quod Erat Demonstrandum

## 2008/11/27

### 證明不等式的基礎招式 (Part 1)

Filed under: Pure Mathematics,Teaching — johnmayhk @ 6:36 下午
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$A - B \ge 0$，或

$\frac{A}{B} \ge 1$ （當知道 $B > 0$ 時）

e.g. 1 For $a, b > 0$, show that $a^ab^b \ge a^bb^a$

$a \ge b$，寫 $\frac{a^ab^b}{a^bb^a} = (\frac{a}{b})^{a-b} \ge 1$
$b \ge a$，寫 $\frac{a^ab^b}{a^bb^a} = (\frac{b}{a})^{b-a} \ge 1$

e.g. 2 For any $p, q, a, b > 0$, show that $a^pb^p - a^pb^q - a^qb^p + a^qb^q \ge 0$

L.H.S. = $(a^p - a^q)(b^p - b^q)$

$p \ge q$ 時，$(a^p - a^q) \ge 0$ and $(b^p - b^q) \ge 0$，故 $(a^p - a^q)(b^p - b^q) \ge 0$
$p \le q$ 時，$(a^p - a^q) \le 0$ and $(b^p - b^q) \le 0$，故 $(a^p - a^q)(b^p - b^q) \ge 0$

$(a^p - a^q)(b^p - b^q) \ge 0$

e.g. 2 的結果是

$a^pb^p + a^qb^q \ge a^pb^q + a^qb^q$

$(a^p - b^p)(a^q - b^q)$，同理，我們可得出不一樣的結果：

$a^{p + q} + b^{p + q} \ge a^pb^q + a^qb^p$

$a_k^{p + q} + b_j^{p + q} \ge a_k^pb_j^q + a_k^qb_j^p$
$\Rightarrow \sum_{k = 1}^{n}\sum_{j = 1}^{n}(a_k^{p + q} + b_j^{p + q}) \ge \sum_{k = 1}^{n}\sum_{j = 1}^{n}(a_k^pb_j^q + a_k^qb_j^p)$
$\Rightarrow n\sum_{k = 1}^{n}a_k^{p + q} + n\sum_{j = 1}^{n}b_j^{p + q} \ge \sum_{k = 1}^{n}a_k^p\sum_{j = 1}^{n}b_j^q + \sum_{k = 1}^{n}a_k^q\sum_{j = 1}^{n}b_j^p$

$n\sum_{k = 1}^{n}a_k^{p + q} \ge \sum_{k = 1}^{n}a_k^p\sum_{k = 1}^{n}a_k^q$

e.g. 3 不少結果都是利用節節相消產生出來，嗯，承上例，設 $p = m, q = 1$ 又假設所有 $a_k$ 是正數，得

$\frac{\sum_{k = 1}^{n}a_k^{m + 1}}{\sum_{k = 1}^{n}a_k^{m}} \ge \frac{1}{n}\sum_{k = 1}^{n}a_k$

$\frac{\sum_{k = 1}^{n}a_k^{2}}{\sum_{k = 1}^{n}a_k^{1}} \ge \frac{1}{n}\sum_{k = 1}^{n}a_k$

$\frac{\sum_{k = 1}^{n}a_k^{3}}{\sum_{k = 1}^{n}a_k^{2}} \ge \frac{1}{n}\sum_{k = 1}^{n}a_k$

$\frac{\sum_{k = 1}^{n}a_k^{4}}{\sum_{k = 1}^{n}a_k^{3}} \ge \frac{1}{n}\sum_{k = 1}^{n}a_k$

$\frac{\sum_{k = 1}^{n}a_k^{m}}{\sum_{k = 1}^{n}a_k^{m - 1}} \ge \frac{1}{n}\sum_{k = 1}^{n}a_k$

$\frac{\sum_{k = 1}^{n}a_k^{m}}{\sum_{k = 1}^{n}a_k^{1}} \ge (\frac{1}{n}\sum_{k = 1}^{n}a_k)^{m - 1}$
$(\frac{1}{n}\sum_{k = 1}^{n}a_k^{m})^{\frac{1}{m}} \ge \frac{1}{n}\sum_{k = 1}^{n}a_k$

$r.m.s. \ge A.M.$

e.g. 4 Given that $a_{k + 1} + b_k \ge a_k + b_{k + 1}$ for $i = 1, 2, \dots n - 1$, show that $a_n + b_1 \ge a_1 + b_n$

$a_2 - a_1 \ge b_2 - b_1$
$a_3 - a_2 \ge b_3 - b_2$
$a_4 - a_3 \ge b_4 - b_3$

$a_{n} - a_{n - 1} \ge b_{n} - b_{n - 1}$

$a_{n} - a_1 \ge b_{n} - b_1$

result follows.

e.g. 5 If $x \in [0.36 , 0.4]$, show that $\frac{28x}{(x^2 + 4)^2} < 0.7$

$\frac{28x}{(x^2 + 4)^2} < \frac{28(0.4)}{(0.36^2 + 4)^2} < 0.7$

e.g. 6 Show that $\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots + \frac{1}{n!} < 2$ for any $n \in \mathbb{N}$.

$\frac{1}{3!} = \frac{1}{2 \times 3} < \frac{1}{2 \times 2} = \frac{1}{2^2}$
$\frac{1}{4!} = \frac{1}{2 \times 3 \times 4} < \frac{1}{2 \times 2 \times 2} = \frac{1}{2^3}$
$\frac{1}{5!} = \frac{1}{2 \times 3 \times 4 \times 5} < \frac{1}{2 \times 2 \times 2 \times 2} = \frac{1}{2^4}$
$\frac{1}{6!} = \frac{1}{2 \times 3 \times 4 \times 5 \times 6} < \frac{1}{2 \times 2 \times 2 \times 2 \times 2} = \frac{1}{2^5}$

$\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots + \frac{1}{n!}$
$< 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^{n - 1}}$
$< 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots$ (sum to infinity)
$= \frac{1}{1 - 1/2}$ (sum of G.S. to infinity)
$= 2$

e.g. 7 For any non-negative numbers $a, b$, show that $\frac{a + b}{2} \ge 2\sqrt{ab}$.

$(\sqrt{a} - \sqrt{b})^2 \ge 0$

e.g. 8 Let $a_1, a_2, \dots a_n; b_1, b_2, \dots b_n$ be $2n$ positive numbers. Show that $(\sum_{k=1}^na_k^2)(\sum_{k=1}^nb_k^2) \ge (\sum_{k=1}^na_kb_k)^2$

$(a_kx + b_k)^2 \ge 0$　　$\forall x \in \mathbb{R}$
$a_k^2x^2 + 2a_kb_kx + b_k^2 \ge 0$　　$\forall x \in \mathbb{R}$
$\sum_{k = 1}^n(a_k^2x^2 + 2a_kb_kx + b_k^2) \ge 0$　　$\forall x \in \mathbb{R}$
$(\sum_{k = 1}^na_k^2)x^2 + 2(\sum_{k = 1}^na_kb_k)x + \sum_{k = 1}^nb_k^2 \ge 0$　　$\forall x \in \mathbb{R}$

Since the above quadratic inequality holds FOR ALL REAL NUMBERS $x$, hence

$\Delta \le 0$

$\Rightarrow (2\sum_{k = 1}^na_kb_k)^2 - 4(\sum_{k = 1}^na_k^2)(\sum_{k = 1}^nb_k^2) \le 0$

Result follows.

$f(x)$ 的最小值（minimum value）是 $a$

e.g. 9 Define $f(x) = a_1 + a_2 + x - 3\sqrt[3]{a_1a_2x}$, where $a_1, a_2 > 0$ with $x \ge 0$. Show that $a_1 + a_2 + a_3 \ge 3\sqrt[3]{a_1a_2a_3}$.
(Note: You may use the result in e.g. 7, but not $A.M. \ge G.M.$)

$f'(x) = 1 - \sqrt[3]{a_1a_2}x^{-\frac{2}{3}}$

Set $f'(x) = 0$, yield

$x = \sqrt{a_1a_2}$ (the unique solution for $x \ge 0$)

Check $f''(a_1a_2) > 0$, hence

$f(x)$ attains its (global) minimum at $x = \sqrt{a_1a_2}$

i.e. $f(x) \ge f(\sqrt{a_1a_2})$　　$\forall x \ge 0$

Just take $x = a_3$, we have

$f(a_3) \ge f(\sqrt{a_1a_2})$
$\Rightarrow a_1 + a_2 + a_3 - 3\sqrt[3]{a_1a_2a_3} \ge a_1 + a_2 + \sqrt{a_1a_2} - 3\sqrt[3]{a_1a_2\sqrt{a_1a_2}}$
$\Rightarrow a_1 + a_2 + a_3 - 3\sqrt[3]{a_1a_2a_3} \ge a_1 + a_2 - 2\sqrt{a_1a_2}$
$\Rightarrow a_1 + a_2 + a_3 - 3\sqrt[3]{a_1a_2a_3} \ge 0$　(by e.g. 7)
$\Rightarrow a_1 + a_2 + a_3 \ge 3\sqrt[3]{a_1a_2a_3}$

e.g. 10 For any real numbers $a, b$, show that $\frac{|a|}{1 + |a|} + \frac{|b|}{1 + |b|} \ge \frac{|a + b|}{1 + |a + b|}$.

$f(x) = \frac{x}{1 + x}$ ; $x \ne -1$

$f'(x) = \frac{1}{(1 + x)^2} > 0$，即 $f(x)$ 是嚴格遞增函數（strictly increasing function），意指 $a > b \Rightarrow f(a) > f(b)$

$\frac{|a|}{1 + |a|} + \frac{|b|}{1 + |b|}$
$= \frac{|a| + 2|ab| + |b|}{1 + |a| + |b| + |ab|}$
$\ge \frac{|a| + |ab| + |b|}{1 + |a| + |b| + |ab|}$
$= f(|a| + |b| + |ab|)$
$\ge f(|a| + |b|)$ ($\because f(x)$ is strictly increasing)
$\ge f(|a + b|)$ ($\because |a| + |b| \ge |a + b|$ and $f(x)$ is strictly increasing)
$= \frac{|a + b|}{1 + |a + b|}$

## 3 則迴響 »

1. 原來這些只是基礎…….

但對我來說，考AL pure equality那part已經夠用了。

迴響 由 Wong Hon — 2008/12/02 @ 1:59 下午 | 回應

2. Hi John,

Very well-written with suitable examples to illustrate the problem-solving strategy. I’ve learnt a great deal even though I’m a teacher myself. Keep writing great articles to benefit all those who are interested in mathematics! Jia you!

Cheers,
Wen Shih

迴響 由 Wen Shih — 2008/12/08 @ 10:15 下午 | 回應

3. It’s very kind of you Wen Shih! What a sweet and encouraging comment from you! I’ll write something (not great articles) if I have time. Thank you again.

迴響 由 johnmayhk — 2008/12/09 @ 8:11 上午 | 回應