418531 ภาคต้น 2552/โจทย์ปัญหาการพิสูจน์ II/เฉลยข้อ 6

ข้อย่อย 1

ก่อนอื่นเราจะนิยามฟังก์ชัน min และ max สำหรับค่าสองค่าก่อน ดังนี้

$\mathrm {min2} (a,b)={\begin{cases}a,&a\leq b\\b,&a>b\end{cases}}$

และ

$\mathrm {max2} (a,b)={\begin{cases}a,&a>b\\b,&a\leq b\end{cases}}$

จากนั้นเราจึงนิยามฟังก์ชัน min และ max สำหรับค่า n ค่าใดๆ ดังต่อไปนี้

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$\min(a_{1},a_{2},\ldots ,a_{n})={\begin{cases}a_{1},&n=1\\\mathrm {min2} (\min(a_{1},a_{2},\ldots ,a_{n-1}),a_{n}),&n>1\end{cases}}$

และ

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$\max(a_{1},a_{2},\ldots ,a_{n})={\begin{cases}a_{1},&n=1\\\mathrm {max2} (\max(a_{1},a_{2},\ldots ,a_{n-1}),a_{n}),&n>1\end{cases}}$

ข้อย่อย 2

ก่อนเราจะพิสูจน์ข้อความในโจทยฺ์ เราจะพิสูจน์ lemma ต่อไปนี้ก่อน

lemma 1: ให้

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$a\,$ และ $b\,$ เป็นจำนวนจริงใดๆ แล้ว

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$\mathrm {max2} (-a,-b)=-\mathrm {min2} (a,b)\,$

พิสูจน์ (lemma 1): การพิสูจน์แบ่งออกเป็นสองกรณี

1. 

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   $a>b\,$ เราได้ว่า
   

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   $-a<-b\,$ ดังนั้น
   

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   $\mathrm {max2} (-a,-b)=-b=-\mathrm {min2} (a,b)\,$
2. $a\leq b\,$ เราได้ว่า
   

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   $-a\geq -b\,$ ดังนั้น
   

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   $\mathrm {max2} (-a,-b)=-a=-\mathrm {min2} (a,b)\,$

พิสูจน์ (โจทย์): เราจะทำการพิสูจน์โดยใช้ induction บนตัวแปร $n\,$

(Base Case) $n=1\,$ เราจะได้ว่า

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$\max(-a_{1})=-a_{1}=-\min(a_{1})\,$

(Induction Case) ให้ n เป็นจำนวนเต็มบวกและสมมติให้ $\max(-a_{1},-a_{2},\ldots ,-a_{n})=-\min(a_{1},a_{2},\ldots ,a_{n})$

ให้

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$a_{1},a_{2},\ldots ,a_{n+1}$ เป็นจำนวนจริงใดๆ เราได้ว่า

	$=\,$
		Error Too many requests (f061ab2) If you report this error to the Wikimedia System Administrators, please include the details below. Request served via cp5021 cp5021, Varnish XID 921848253 Upstream caches: cp5021 int Error: 429, Too many requests (f061ab2) at Sun, 15 Feb 2026 21:27:37 GMT Sensitive client information IP address: 158.108.32.49 $-\mathrm {min2} (\min(a_{1},a_{2},\ldots ,a_{n}),a_{n+1})\,$	(ใช้ lemma 1)
		$-\min(a_{1},a_{2},\ldots ,a_{n+1})\,$

ดังนั้นเราสามารถสรุปได้ว่า สำหรับจำนวนจริง ใดๆ

ข้อย่อย 3

ก่อนเราจะพิสูจน์ข้อความในโจทยฺ์ เราจะพิสูจน์ lemma ต่อไปนี้ก่อน

lemma 2: ให้

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$a_{1},a_{2}\,$ และ

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$b_{1},b_{2}\,$ เป็นจำนวนจริงใดๆ แล้ว

พิสูจน์ (lemma 2): ให้

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$A=\mathrm {max2} (a_{1},a_{2})\,$ และให้ เราได้ว่า และ ฉะนั้น และ

เนื่องจาก

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$\mathrm {max2} (a_{1}+b_{1},a_{2}+b_{2})\,$ มีค่าเท่ากับ หรือไม่ก็

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$a_{2}+b_{2}\,$ เราจึงได้ว่า

พิสูจน์ (โจทย์): เราจะทำการพิสูจน์โดยใช้ induction บนตัวแปร

(Base Case) เราจะได้ว่า

(Induction Case) ให้ n เป็นจำนวนเต็มบวกและสมมติให้ $\max(a_{1}+b_{1},a_{2}+b_{2},\ldots ,a_{n}+b_{n})\leq \max(a_{1},a_{2},\ldots ,a_{n})+\max(b_{1},b_{2},\ldots ,b_{n})\,$

ให้

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$a_{1},a_{2},\ldots ,a_{n+1},b_{1},b_{2},\ldots ,b_{n+1}$ เป็นจำนวนจริงใดๆ เราได้ว่า

	$=\,$
			(สมมติฐาน)
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	$=\,$

ดังนั้นเราสามารถสรุปได้ว่า $\max(a_{1}+b_{1},a_{2}+b_{2},\ldots ,a_{n}+b_{n})\leq \max(a_{1},a_{2},\ldots ,a_{n})+\max(b_{1},b_{2},\ldots ,b_{n})$ สำหรับจำนวนจริง $a_{1},a_{2},\ldots ,a_{n}$ ใดๆ

ข้อย่อย 4

ก่อนเราจะพิสูจน์ข้อความในโจทยฺ์ เราจะพิสูจน์ lemma ต่อไปนี้ก่อน

lemma 3: ให้ และ

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$b_{1},b_{2}\,$ เป็นจำนวนจริงใดๆ แล้ว

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$\mathrm {min2} (a_{1}+b_{1},a_{2}+b_{2})\geq \mathrm {min2} (a_{1},a_{2})+\mathrm {min2} (b_{1},b_{2})\,$

พิสูจน์ (lemma 3): ให้ และให้ เราได้ว่า และ $b_{2}\geq \beta \,$ ฉะนั้น และ

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$a_{2}+b_{2}\geq \alpha +\beta \,$

เนื่องจาก

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$\mathrm {min2} (a_{1}+b_{1},a_{2}+b_{2})\,$ มีค่าเท่ากับ หรือไม่ก็

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$a_{2}+b_{2}\,$ เราจึงได้ว่า

พิสูจน์ (โจทย์): เราจะทำการพิสูจน์โดยใช้ induction บนตัวแปร $n\,$

(Base Case) เราจะได้ว่า

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$\min(a_{1}+b_{1})\leq a_{1}+b_{1}=\min(a_{1})+\min(b_{1})\,$

(Induction Case) ให้ n เป็นจำนวนเต็มบวกและสมมติให้

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$\min(a_{1}+b_{1},a_{2}+b_{2},\ldots ,a_{n}+b_{n})\leq \min(a_{1},a_{2},\ldots ,a_{n})+\min(b_{1},b_{2},\ldots ,b_{n})\,$

ให้ เป็นจำนวนจริงใดๆ เราได้ว่า

		Error Too many requests (f061ab2) If you report this error to the Wikimedia System Administrators, please include the details below. Request served via cp5021 cp5021, Varnish XID 960007415 Upstream caches: cp5021 int Error: 429, Too many requests (f061ab2) at Sun, 15 Feb 2026 21:27:43 GMT Sensitive client information IP address: 158.108.32.49 $\mathrm {min2} (\min(a_{1},a_{2},\ldots ,a_{n})+\min(b_{1},b_{2},\ldots ,b_{n}),a_{n+1}+b_{n+1})\,$	(สมมติฐาน)
	$\geq \,$	Error Too many requests (f061ab2) If you report this error to the Wikimedia System Administrators, please include the details below. Request served via cp5021 cp5021, Varnish XID 963970542 Upstream caches: cp5021 int Error: 429, Too many requests (f061ab2) at Sun, 15 Feb 2026 21:27:44 GMT Sensitive client information IP address: 158.108.32.49 $\mathrm {min2} (\min(a_{1},a_{2},\ldots ,a_{n}),a_{n+1})+\mathrm {min2} (\min(b_{1},b_{2},\ldots ,b_{n}),b_{n+1})\,$	(ใช้ lemma 3)

ดังนั้นเราสามารถสรุปได้ว่า สำหรับจำนวนจริง $a_{1},a_{2},\ldots ,a_{n}$ ใดๆ