Nicolae Paun 2008 Problema 3

[maxim bogdan]

Sa se arate ca: \( 1+x+x^2+x^3+\dots+x^{2n}\geq\frac{n+1}{2n+1},(\forall)n\in\mathbb{N}* \), si pentru orice \( x \) real.

Sorin Radulescu, Bucuresti

[maxim bogdan]

Daca \( x\geq 0 \), atunci \( 1+x+x^2+\dots+x^{2n}>1>\frac{n+1}{2n+1}. \)

Daca \( x\leq -1 \) atunci \( 1+x+x^2+\dots+x^{2n}=1+(x+x^2)+(x^3+x^4)+\dots+(x^{2n-1}+x^{2n})>1>\frac{n+1}{2n+1}, \) deoarece \( x^{2n-1}<0<x^{2n}. \)

Sa demonstram inegalitatea pentru \( x\in(-1;0). \)

Avem: \( 1+x+x^2+\dots+x^{2n}=\frac{1-x^{2n+1}}{1-x}\geq \frac{n+1}{2n+1}\Longleftrightarrow \)

\( \Longleftrightarrow (2n+1)(-x)^{2n+1}+n\geq (n+1)(-x). \) Fie \( a=-x \), deci \( a\in(0;1). \)

Inegalitatea de demonstat: \( (2n+1)a^{2n+1}+n\geq (n+1)a \)

\( (2n+1)a^{2n+1}+n>n\geq(n+1)a\Longleftrightarrow a\leq\frac{n}{n+1} \).


Mai avem de demonstrat inegalitatea daca \( a>\frac{n}{n+1}(*) \).


\( (2n+1)a^{2n+1}+n>(2n+1)\cdot(\frac{n}{n+1})^n\cdot a^{n+1}+n=(2n+1)\cdot\frac{1}{(1+\frac{1}{n})^n}\cdot a^{n+1}+n>\frac{2n+1}{3}\cdot a^{n+1}+n\geq a^{n+1}+n=a^{n+1}+1+1\dots+1\geq (n+1)\sqrt[n+1]{a^{n+1}}=(n+1)a. \)

Prima data am aplicat inegalitatea \( (*) \), a doua oara am folosit cunoscuta inegalitate \( (1+\frac{1}{n})^n<e=2,71...<3 \)(demonstratie prin inductie), iar a treia oara inegalitatea mediilor.

Deci problema este rezolvata!

[Marius Mainea]

Solutia 2: Inductie dupa n.

[Claudiu Mindrila]

Intr-adevar, vom arata implicatia \( P\left(n\right)\longrightarrow P\left(n+1\right) \). Insa vom arata mai intai ca \( P\left(1\right):\ 1+x+x^{2}\ge\frac{3}{5} \) este adevarata. Intr-adevar \( \ 1+x+x^{2}\ge\frac{3}{5}\Longleftrightarrow f\left(x\right)=x^{2}+x+\frac{2}{5}\ge0 \). Cum \( \ 1+x+x^{2}\ge\frac{3}{5}\Longleftrightarrow f\left(x\right)=x^{2}+x+\frac{2}{5}\ge0 \) si \( \triangle_{f}=1-4\cdot\frac{2}{5}=-\frac{3}{5}<0\Longrightarrow f\left(x\right)>0 \).
Acum, implicatia \( P\left(n\right)\longrightarrow P\left(n+1\right) \) revine la \( \sum_{i=0}^{2n}x^{i}>\frac{n+1}{2n+1}\Longrightarrow\sum_{i=0}^{2n+2}x^{i}>\frac{n+2}{2n+3} \). Dar \( \sum_{i=0}^{2n+2}=1+x+x^{2}\left(\sum_{i=0}^{2n}x^{i}\right)>1+x+x^{2}\cdot\frac{n+1}{2n+1} \). Mai ramane acum de aratat ca \( 1+x+x^{2}\cdot\frac{n+1}{2n+1}>\frac{n+2}{2n+3}\Longleftrightarrow g\left(x\right)=x^{2}\cdot\frac{n+1}{2n+1}+x+\frac{n+1}{2n+3}>0 \), dar cum \( \triangle_{g}=1-\frac{4\left(n+1\right)^{2}}{\left(2n+1\right)\left(2n+3\right)}<0\Longleftrightarrow\left(2n+1\right)\left(2n+3\right)<4\left(n+1\right)^{2} \) problema este rezolvata. Ultima inegalitate rezulta din inegalitatea mediilor: \( \left(2n+1\right)\left(2n+3\right)<\frac{\left(2n+1+2n+3\right)^{2}}{4}=4\left(n+1\right)^{2} \)