Quod Erat Demonstrandum

十一月 27, 2008

證明不等式的基礎招式 (Part 1)

Filed under: Pure Mathematics, Teaching — johnmayhk @ 6:36 pm
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證明不等式的招式繁多，難以窮盡，聊舉數例，全屬基礎技術，以作溫習之用，高手見諒。

招式一．基本功

欲證 A \ge B

可試證：

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

可試證 \frac{a^ab^b}{a^bb^a} \ge 1

為方便看：

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

招式二．分 cases

其實 e.g. 1 也有一些分 cases 的情況，再舉一例：

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)

當 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
當 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

綜合上述兩個結果，得

(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

如果考慮

(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

大家試試。把上述結果繼續引申，可以得到某類排序不等式，見下

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

進一步設：b_i = a_i，上式即

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

招式三．節節相消

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

依次代入不同的正整數 m，可得

\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

把上述 m - 1 條不等式相乘，得：

\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

特別地，如果取 m = 2，上式就變成我們熟悉的

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

不過是普通的節節相消，依次代入 k = 1, 2, \dots , n - 1，得

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}

把上述 n - 1 條不等式相加，得：

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

result follows.

招式四．改變分子分母

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

既知 x \in [0.36 , 0.4]，我們把 \frac{28x}{(x^2 + 4)^2} 的分子「作大」，即分子最大可能都不過是 28(0.4)；同樣地，我們把分母「作細」，即 \frac{28x}{(x^2 + 4)^2} 的分母最小可能不過是 (0.36^2 + 4)^2，於是

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

當然，大家也可用求導法處理。

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}.

把 L.H.S. 每項的分母「作細」，得到所謂的「上限」，見下：

\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}

於是，

\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

招式五：平方非負

無論是證明 A.M. \ge G.M. 或 P.S. \ge S.P. (product of square \ge square of product, 即 C.S. inequality 是也)，起手式都是考慮：平方非負。

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

以 m.i. 來證明 A.M. \ge G.M.，通常就以 e.g. 7 的結果做起。考慮平方非負，即

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

化簡後立即得：\frac{a + b}{2} \ge 2\sqrt{ab}

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

這其實是 C.S. inequality 的簡單版，證明的開端也是考慮：平方非負，見下：

(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.

招式六：求導法（Differentiation）

觀察：f(x) \ge a for all x \in \mathbb{R}（Given that f(x_0) = a for some x_0）

以另一個表達上述不等式，可以說：

f(x) 的最小值（minimum value）是 a。

尋求最小值或最大值，可以利用求導法。

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|}.

這例用到三角形不等式（Triangle inequality）：|a| + |b| \ge |a + b|

設 f(x) = \frac{x}{1 + x} ; x \ne -1

因 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)。

現在，

\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|}

待續

3個回應 »

  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。

    Comment by Wong Hon — 十二月 2, 2008 @ 1:59 pm | Reply

  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

    Comment by Wen Shih — 十二月 8, 2008 @ 10:15 pm | Reply

  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.

    Comment by johnmayhk — 十二月 9, 2008 @ 8:11 am | Reply


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