Točno
23. lipnja 2016. 00:58 (8 godine)
Upozorenje: Ovaj zadatak još niste riješili!
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Kliknite ovdje kako biste prikazali rješenje.
Pokušat ćemo tvrdnju zadatka dokazati Cauchyevom indukcijom:
Baza:![n = 2](/media/m/5/d/e/5de286c03f7be13a5a67f0265685173b.png)
Imamo:
![\frac{1}{r_1+1} + \frac{1}{r_2+1} \geq \frac{2}{\sqrt{r_1r_2}+1}](/media/m/b/1/8/b18db43e900953941e7583bb5aa846bb.png)
ali lako se možemo uvjeriti (oduzimanjem desne strane od lijeve) da:
![\frac{1}{r_1+1} + \frac{1}{r_2+1} = \frac{2}{\sqrt{r_1r_2}+1} + \frac{(\sqrt{r_1}-\sqrt{r_2})^2(\sqrt{r_1r_2}-1)}{(r_1+1)(r_2+1)(\sqrt{r_1r_2}+1)}](/media/m/7/1/1/711d8f1b3b8a1fe1e66b41f96885f642.png)
što očito dokazuje gore navedenu nejednakost za![r_1, r_2 \geq 1](/media/m/d/d/3/dd3ac6dec66fa3dfcb3b9b3d14b66518.png)
Pretpostavka: za neki
vrijedi:
![\sum_{i=1}^{n} \frac{1}{r_{i} + 1} \geq \frac{n}{ \sqrt[n]{r_{1} \cdots r_{n}}+1}](/media/m/9/2/a/92adfde36c366097f34bfc4eb78ed089.png)
Korak:![n \rightarrow 2n](/media/m/1/f/3/1f3df244481c8b39fab86cc0b5dcf00e.png)
Imamo:
![\sum_{i=1}^{2n} \frac{1}{r_{i} + 1} = \sum_{i=1}^{n} \frac{1}{r_{i} + 1} + \sum_{i=n+1}^{2n} \frac{1}{r_{i} + 1} \geq n \cdot\Bigg(\frac{1}{ \sqrt[n]{r_{1} \cdots r_{n}}+1}+\frac{1}{ \sqrt[n]{r_{n+1} \cdots r_{2n}}+1}\Bigg) \geq \frac{2n}{\sqrt[2n]{r_1\cdots r_{2n}}+1}](/media/m/a/9/9/a99cc9359ec55c155452c0ccc2693425.png)
Čime je korak dokazan.
Korak:![n \rightarrow n-1](/media/m/e/a/b/eab4735a877eec9d2240a10eaa845cd0.png)
Uzmimo:![r_n = \sqrt[n-1]{r_1\cdots r_{n-1}}](/media/m/6/a/a/6aa91d10cad79eb39cada4d8d8234b51.png)
Tada imamo:
![\frac{1}{ \sqrt[n-1]{r_1\cdots r_{n-1}}+1}+\sum_{i=1}^{n-1} \frac{1}{r_{i} + 1} \geq \frac{n}{ (r_{1} \cdots r_{n-1}\cdot (r_{1} \cdots r_{n-1})^\frac{1}{n-1})^{\frac{1}{n}}+1}= \frac{n}{(((r_{1} \cdots r_{n-1})^{\frac{n}{n-1}})^{\frac{1}{n}}+1} = \frac{n-1}{\sqrt[n-1]{r_1\cdots r_{n-1}}+1} + \frac{1}{\sqrt[n-1]{r_1\cdots r_{n-1}}+1}](/media/m/f/d/f/fdf754a6a1bea50c37b0b6e6945903aa.png)
![\iff](/media/m/0/1/9/0195c4069e7f949cd8f011d602abaf24.png)
![\sum_{i=1}^{n-1} \frac{1}{r_{i} + 1} \geq \frac{n-1}{\sqrt[n-1]{r_1\cdots r_{n-1}}+1}](/media/m/7/f/d/7fdc56218dd193882ee93116586ae741.png)
Čime je korak dokazan.
Po pretpostavci matematičke indukcije, tvrdnja zadatka vrijedi za
.
Baza:
![n = 2](/media/m/5/d/e/5de286c03f7be13a5a67f0265685173b.png)
Imamo:
![\frac{1}{r_1+1} + \frac{1}{r_2+1} \geq \frac{2}{\sqrt{r_1r_2}+1}](/media/m/b/1/8/b18db43e900953941e7583bb5aa846bb.png)
ali lako se možemo uvjeriti (oduzimanjem desne strane od lijeve) da:
![\frac{1}{r_1+1} + \frac{1}{r_2+1} = \frac{2}{\sqrt{r_1r_2}+1} + \frac{(\sqrt{r_1}-\sqrt{r_2})^2(\sqrt{r_1r_2}-1)}{(r_1+1)(r_2+1)(\sqrt{r_1r_2}+1)}](/media/m/7/1/1/711d8f1b3b8a1fe1e66b41f96885f642.png)
što očito dokazuje gore navedenu nejednakost za
![r_1, r_2 \geq 1](/media/m/d/d/3/dd3ac6dec66fa3dfcb3b9b3d14b66518.png)
Pretpostavka: za neki
![n \in \mathbb{N}](/media/m/2/b/a/2ba27c6141ca415bb86bae1d237f1fac.png)
![\sum_{i=1}^{n} \frac{1}{r_{i} + 1} \geq \frac{n}{ \sqrt[n]{r_{1} \cdots r_{n}}+1}](/media/m/9/2/a/92adfde36c366097f34bfc4eb78ed089.png)
Korak:
![n \rightarrow 2n](/media/m/1/f/3/1f3df244481c8b39fab86cc0b5dcf00e.png)
Imamo:
![\sum_{i=1}^{2n} \frac{1}{r_{i} + 1} = \sum_{i=1}^{n} \frac{1}{r_{i} + 1} + \sum_{i=n+1}^{2n} \frac{1}{r_{i} + 1} \geq n \cdot\Bigg(\frac{1}{ \sqrt[n]{r_{1} \cdots r_{n}}+1}+\frac{1}{ \sqrt[n]{r_{n+1} \cdots r_{2n}}+1}\Bigg) \geq \frac{2n}{\sqrt[2n]{r_1\cdots r_{2n}}+1}](/media/m/a/9/9/a99cc9359ec55c155452c0ccc2693425.png)
Čime je korak dokazan.
Korak:
![n \rightarrow n-1](/media/m/e/a/b/eab4735a877eec9d2240a10eaa845cd0.png)
Uzmimo:
![r_n = \sqrt[n-1]{r_1\cdots r_{n-1}}](/media/m/6/a/a/6aa91d10cad79eb39cada4d8d8234b51.png)
Tada imamo:
![\frac{1}{ \sqrt[n-1]{r_1\cdots r_{n-1}}+1}+\sum_{i=1}^{n-1} \frac{1}{r_{i} + 1} \geq \frac{n}{ (r_{1} \cdots r_{n-1}\cdot (r_{1} \cdots r_{n-1})^\frac{1}{n-1})^{\frac{1}{n}}+1}= \frac{n}{(((r_{1} \cdots r_{n-1})^{\frac{n}{n-1}})^{\frac{1}{n}}+1} = \frac{n-1}{\sqrt[n-1]{r_1\cdots r_{n-1}}+1} + \frac{1}{\sqrt[n-1]{r_1\cdots r_{n-1}}+1}](/media/m/f/d/f/fdf754a6a1bea50c37b0b6e6945903aa.png)
![\iff](/media/m/0/1/9/0195c4069e7f949cd8f011d602abaf24.png)
![\sum_{i=1}^{n-1} \frac{1}{r_{i} + 1} \geq \frac{n-1}{\sqrt[n-1]{r_1\cdots r_{n-1}}+1}](/media/m/7/f/d/7fdc56218dd193882ee93116586ae741.png)
Čime je korak dokazan.
Po pretpostavci matematičke indukcije, tvrdnja zadatka vrijedi za
![\forall n\in \mathbb{N}](/media/m/2/a/f/2afc9c816ab6781c829d2aaa18bd8df5.png)