27401 - Revision history

Andrei.Horvat at 08:36, 1 December 2024

2024-12-01T08:36:43Z

← Older revision		Revision as of 08:36, 1 December 2024
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	'''27401 (Radu Pop)'''		'''27401 (Radu Pop)'''

Andrei.Horvat at 10:47, 12 January 2024

2024-01-12T10:47:41Z

← Older revision		Revision as of 10:47, 12 January 2024
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	'''27401 (Radu Pop~~, Baia Mare~~)'''		'''27401 (Radu Pop)'''

	''Fie <math>n \in \mathbb{N}</math>. Să se arate că		''Fie <math>n \in \mathbb{N}</math>. Să se arate că
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	'''Soluție:'''		'''Soluție:'''

	Fie <math> x,y \in [0,\infty)</math>.Avem		Fie <math> x,y \in [0,\infty)</math>. Avem
	<math display="block">(x+1)(y+1)(x+y+n^2+n)=(n+1) \biggl(\underbrace{ \frac{x}{n + 1}+ \frac{x}{n + 1}+ \ldots +\frac{x}{n + 1}}_{(n+1)\text{ ori}}+1 \biggr) \cdot </math>		<math display="block">(x+1)(y+1)(x+y+n^2+n)=(n+1) \biggl(\underbrace{ \frac{x}{n + 1}+ \frac{x}{n + 1}+ \ldots +\frac{x}{n + 1}}_{(n+1)\text{ ori}}+1 \biggr) \cdot </math>

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	<math display="block">=\frac{xy}{n+1}(n+2)^3.</math>		<math display="block">=\frac{xy}{n+1}(n+2)^3.</math>

	Rezultă că <math>(n+1)(x+1)(y+1)(x+y+n^2+n) \ge (n+2)^3xy</math>. Fie <math>x=a-1\ge 0</math> şi <math>y=b-1\ge0</math>. Obţinem <math>(n+1)ab(a+b+n^2+n-2)\ge(n+2)^3(a-1)(b-1)</math>, de unde <math>(n+1)ab(a+b)+(n^3+2n^2-n-2)ab \ge (n+2)^3ab-(n+2)^3(a+b)+(n+2)^3</math>, deci <math>(n+1)ab(a+b)-(4n^2+13n+10)ab+(n+2)^3(a+b) \ge (n+2)^3</math>.		Rezultă că <math display="block">(n+1)(x+1)(y+1)(x+y+n^2+n) \ge (n+2)^3xy.</math>Fie <math>x=a-1\ge 0</math> şi <math>y=b-1\ge0</math>.

			Obţinem <math display="block">(n+1)ab(a+b+n^2+n-2)\ge(n+2)^3(a-1)(b-1),</math>de unde <math display="block">(n+1)ab(a+b)+(n^3+2n^2-n-2)ab \ge (n+2)^3ab-(n+2)^3(a+b)+(n+2)^3</math>deci <math display="block">(n+1)ab(a+b)-(4n^2+13n+10)ab+(n+2)^3(a+b) \ge (n+2)^3.</math>

Fellner Arthur at 20:55, 11 January 2024

2024-01-11T20:55:22Z

← Older revision		Revision as of 20:55, 11 January 2024
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	<math display="block">\ge (n+1)\sqrt[n+2]{\frac{x^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{y^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{xy}{(n+1)^{2}}} \cdot (n+2)^3 =</math>		<math display="block">\ge (n+1)\sqrt[n+2]{\frac{x^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{y^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{xy}{(n+1)^{2}}} \cdot (n+2)^3 =</math>

	<math display="block">\frac{xy}{n+1}(n+2)^3.</math>		<math display="block">=\frac{xy}{n+1}(n+2)^3.</math>

	Rezultă că <math>(n+1)(x+1)(y+1)(x+y+n^2+n) \ge (n+2)^3xy</math>. Fie <math>x=a-1\ge 0</math> şi <math>y=b-1\ge0</math>. Obţinem <math>(n+1)ab(a+b+n^2+n-2)\ge(n+2)^3(a-1)(b-1)</math>, de unde <math>(n+1)ab(a+b)+(n^3+2n^2-n-2)ab \ge (n+2)^3ab-(n+2)^3(a+b)+(n+2)^3</math>, deci <math>(n+1)ab(a+b)-(4n^2+13n+10)ab+(n+2)^3(a+b) \ge (n+2)^3</math>.		Rezultă că <math>(n+1)(x+1)(y+1)(x+y+n^2+n) \ge (n+2)^3xy</math>. Fie <math>x=a-1\ge 0</math> şi <math>y=b-1\ge0</math>. Obţinem <math>(n+1)ab(a+b+n^2+n-2)\ge(n+2)^3(a-1)(b-1)</math>, de unde <math>(n+1)ab(a+b)+(n^3+2n^2-n-2)ab \ge (n+2)^3ab-(n+2)^3(a+b)+(n+2)^3</math>, deci <math>(n+1)ab(a+b)-(4n^2+13n+10)ab+(n+2)^3(a+b) \ge (n+2)^3</math>.

Fellner Arthur at 20:34, 11 January 2024

2024-01-11T20:34:48Z

← Older revision		Revision as of 20:34, 11 January 2024
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	~~\usepackage{MnSymbol}~~

	'''27401 (Radu Pop, Baia Mare)'''		'''27401 (Radu Pop, Baia Mare)'''

Fellner Arthur at 20:34, 11 January 2024

2024-01-11T20:34:08Z

← Older revision		Revision as of 20:34, 11 January 2024
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	\biggl(\frac{x}{n+1} + \frac{y}{n+1} +\underbrace{1+1+\cdots + 1 }_{n\text{ ori}}\biggr) \ge</math>		\biggl(\frac{x}{n+1} + \frac{y}{n+1} +\underbrace{1+1+\cdots + 1 }_{n\text{ ori}}\biggr) \ge</math>

	<math display="block">\ge (n+1)\sqrt[n+2]{\frac{x^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{y^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{xy}{(n+1)^{2}}} \cdot (n+2)^3) =</math>		<math display="block">\ge (n+1)\sqrt[n+2]{\frac{x^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{y^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{xy}{(n+1)^{2}}} \cdot (n+2)^3 =</math>

	<math display="block">\frac{xy}{n+1}(n+2)^3.</math>		<math display="block">\frac{xy}{n+1}(n+2)^3.</math>

	Rezultă că <math>(n+1)(x+1)(y+1)(x+y+n^2+n) \ge (n+2)^3xy</math>. Fie <math>x=a-1\ge 0</math> şi <math>y=b-1\ge0</math>. Obţinem <math>(n+1)ab(a+b+n^2+n-2)\ge(n+2)^3(a-1)(b-1)</math>, de unde <math>(n+1)ab(a+b)+(n^3+2n^2-n-2)ab \ge (n+2)^3ab-(n+2)^3(a+b)+(n+2)^3</math>, deci <math>(n+1)ab(a+b)-(4n^2+13n+10)ab+(n+2)^3(a+b) \ge (n+2)^3</math>.		Rezultă că <math>(n+1)(x+1)(y+1)(x+y+n^2+n) \ge (n+2)^3xy</math>. Fie <math>x=a-1\ge 0</math> şi <math>y=b-1\ge0</math>. Obţinem <math>(n+1)ab(a+b+n^2+n-2)\ge(n+2)^3(a-1)(b-1)</math>, de unde <math>(n+1)ab(a+b)+(n^3+2n^2-n-2)ab \ge (n+2)^3ab-(n+2)^3(a+b)+(n+2)^3</math>, deci <math>(n+1)ab(a+b)-(4n^2+13n+10)ab+(n+2)^3(a+b) \ge (n+2)^3</math>.

Fellner Arthur: Pagină nouă: \usepackage{MnSymbol} '''27401 (Radu Pop, Baia Mare)''' ''Fie $n \in \mathbb{N}$ . Să se arate că $(n+1)ab(a+b)-(4n^3+13n+10)ab+(n+2)^3(a+b) \geq (n+2)^3,$ oricare ar fi $a,b \in [1,\infty)$ '' '''Soluție:''' Fie $x,y \in [0,\infty)$ .Avem $(x+1)(y+1)(x+y+n^2+n)=(n+1) \biggl(\underbrace{ \frac{x}{n + 1}+ \frac{x}{n + 1}+ \ldots +\frac{x}{n + 1}}_{(n+1)\text{ ori}}+1 \biggr) \cdot$ ...

2024-01-11T20:31:48Z

Pagină nouă: \usepackage{MnSymbol} '''27401 (Radu Pop, Baia Mare)''' ''Fie <math>n \in \mathbb{N}</math>. Să se arate că <math display="block"> (n+1)ab(a+b)-(4n^3+13n+10)ab+(n+2)^3(a+b) \geq (n+2)^3, </math> oricare ar fi <math>a,b \in [1,\infty)</math>'' '''Soluție:''' Fie <math> x,y \in [0,\infty)</math>.Avem <math display="block">(x+1)(y+1)(x+y+n^2+n)=(n+1) \biggl(\underbrace{ \frac{x}{n + 1}+ \frac{x}{n + 1}+ \ldots +\frac{x}{n + 1}}_{(n+1)\text{ ori}}+1 \biggr) \cdot </math>...

New page

\usepackage{MnSymbol}

'''27401 (Radu Pop, Baia Mare)'''

''Fie <math>n \in \mathbb{N}</math>. Să se arate că
<math display="block"> (n+1)ab(a+b)-(4n^3+13n+10)ab+(n+2)^3(a+b) \geq (n+2)^3, </math>
oricare ar fi <math>a,b \in [1,\infty)</math>''

'''Soluție:'''

Fie <math> x,y \in [0,\infty)</math>.Avem
<math display="block">(x+1)(y+1)(x+y+n^2+n)=(n+1) \biggl(\underbrace{ \frac{x}{n + 1}+ \frac{x}{n + 1}+ \ldots +\frac{x}{n + 1}}_{(n+1)\text{ ori}}+1 \biggr) \cdot </math>

<math display="block"> \cdot \biggl(\underbrace{\frac{y}{n+1} + \frac{y}{n+1} + \cdots + \frac{y}{n+1}}_{(n+1) \text{ ori}} + 1\biggr) \cdot
\biggl(\frac{x}{n+1} + \frac{y}{n+1} +\underbrace{1+1+\cdots + 1 }_{n\text{ ori}}\biggr) \ge</math>

<math display="block">\ge (n+1)\sqrt[n+2]{\frac{x^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{y^{n+1}}{(n+1)^{n+1}}} \cdot \sqrt[n+2]{\frac{xy}{(n+1)^{2}}} \cdot (n+2)^3) =</math>

<math display="block">\frac{xy}{n+1}(n+2)^3.</math>

Rezultă că <math>(n+1)(x+1)(y+1)(x+y+n^2+n) \ge (n+2)^3xy</math>. Fie <math>x=a-1\ge 0</math> şi <math>y=b-1\ge0</math>. Obţinem <math>(n+1)ab(a+b+n^2+n-2)\ge(n+2)^3(a-1)(b-1)</math>, de unde <math>(n+1)ab(a+b)+(n^3+2n^2-n-2)ab \ge (n+2)^3ab-(n+2)^3(a+b)+(n+2)^3</math>, deci <math>(n+1)ab(a+b)-(4n^2+13n+10)ab+(n+2)^3(a+b) \ge (n+2)^3</math>.