Inegalitate conditionata

Claudiu Mindrila · Post by **Claudiu Mindrila** » Mon Jun 09, 2008 9:39 pm

Daca \( a,b,c,d \in[0,1] \) astfel incat \( ab+bc+cd+da=2 \). Demonstrati ca \( a^n+b^n+c^n+d^n \leq 3 \), \( (\forall) n \in \mathbb{N}^* \)

Post by **Laurian Filip** » Mon Jun 09, 2008 9:48 pm

Am si eu o intrebare... Daca arati ca e adevarat pentru n=1 nu e adevarat pentru orice n?

Claudiu Mindrila · Post by **Claudiu Mindrila** » Mon Jun 09, 2008 9:52 pm

Nu cred. \( n=1 \) e doar o valoare. [/tex]

Post by **Laurian Filip** » Mon Jun 09, 2008 9:54 pm

cum a,b,c,d sunt subunitare \( f(n)=a^n+b^n+c^n+d^n \) nu e descrescatoare?

Claudiu Mindrila · Post by **Claudiu Mindrila** » Mon Jun 09, 2008 9:54 pm

La clasa a VII-a

Post by **Laurian Filip** » Mon Jun 09, 2008 9:56 pm

de clasa a 7-a... pentru \( 0<a<1 \) nu e tot timpul \( a>a^n \) pentru orice \( n>1 \)?

Claudiu Mindrila · Post by **Claudiu Mindrila** » Mon Jun 09, 2008 9:59 pm

Ba da... . Ai dreptate Laurian Filip. Mai ramane de aratat ca \( a+b+c+d \leq 3 \)

Marius Mainea · Post by **Marius Mainea** » Mon Jun 09, 2008 10:00 pm

Claudiu, daca demonstram relatia pentru n=1 rezulta si pentru celalte cazuri.

Avem \( \sum (1-a)(1-b)\geq 0 \) de unde \( 4-2\sum a+\sum ab\geq 0 \) si de aici concluzia problemei.

BogdanCNFB · Post by **BogdanCNFB** » Wed Jun 18, 2008 8:02 pm

Avem \( ab+bc+cd+da=(b+d)(a+c) \). Deci, \( (b+d)(a+c)=2 \)\( \Longleftrightarrow\ a+c=\frac{2}{b+d} \)
Aratam in prima faza ca \( a+b+c+d\leq 3\Longleftrightarrow b+d+\frac{2}{b+d}\leq 3\Longleftrightarrow (b+d)^2+2\leq 3(b+d)\Longleftrightarrow (b+d-1)^2\leq b+d-1 \) ceea ce este adevarat pentru ca \( 0\leq b+d-1\leq 1\ \ ;\ \ 0\leq b\leq 1 \) si \( 0\leq d\leq 1 \)
Avem \( a\leq 1\Longrightarrow a^n\leq a \)
\( b\leq 1\Longrightarrow b^n\leq b \)
\( c\leq 1\Longrightarrow c^n\leq c \)
\( d\leq 1\Longrightarrow d^n\leq d \)
Rezulta, prin insumare, ca \( a^n+b^n+c^n+d^n\leq a+b+c+d\leq 3 \)