Fie numarul natural \( n\ge 3 \) si numerele reale pozitive \( x_k\ ,\ k\in\overline {1,n} \) pentru care notam
\( 2s=\sum_{k=1}^nx_k \) . Sa se arate ca \( s\ >\ \max_{1\le k\le n}\ x_k\ \Longrightarrow\ \prod_{k=1}^n\frac {(n-3)s+x_k}{s-x_k}\ \ge\ (n-1)^n \) .
Cazuri particulare.
\( \odot\ (n=3) \) . Intr-un triunghi \( ABC \) exista inegalitatea \( \frac {abc}{(p-a)(p-b)(p-c)}\ \ge\ 8 \) .
\( \odot\ (n=4) \) . Intr-un patrulater convex \( ABCD \) exista inegalitatea \( \prod\ \frac {p+a}{p-a}\ \ge\ 81 \) .
Inegalitate simetrica in n variabile.
Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Notam \( s-x_k=a_k, k=\overline{1,n} \) si de aici prin sumare \( (n-2)s=\sum{a_k} \) sau \( x_1+x_2+...+x_n=\frac{2}{n-2}(a_1+a_2+...+a_n) \) (1)
Scazand din (1) fiecare din egalitatile
\( x1+x_2+...-_k+...+x_n=2a_k , k=\overline{1,n} \) obtinem \( x_k=\sum_{i\neq k}{a_i}, k=\overline{1,n} \) ,
deci inegalitatea este echivalenta cu
\( \prod_{k=1}^n{(a_1+a_2+..+a_{k-1}+a_{k+1}+...+a_n)}\ge (n-1)^na_1\cdot a_2...\cdot a_n \) care rezulta imediat din inegalitatea mediilor,
\( \prod_{k=1}^n{(a_1+a_2+..+a_{k-1}+a_{k+1}+...+a_n)}\ge \prod_{k=1}^n{\sqrt[n-1]{a_1a_2..a_{k-1}a_{k+1}...a_n}= (n-1)^na_1\cdot a_2...\cdot a_n \)
Scazand din (1) fiecare din egalitatile
\( x1+x_2+...-_k+...+x_n=2a_k , k=\overline{1,n} \) obtinem \( x_k=\sum_{i\neq k}{a_i}, k=\overline{1,n} \) ,
deci inegalitatea este echivalenta cu
\( \prod_{k=1}^n{(a_1+a_2+..+a_{k-1}+a_{k+1}+...+a_n)}\ge (n-1)^na_1\cdot a_2...\cdot a_n \) care rezulta imediat din inegalitatea mediilor,
\( \prod_{k=1}^n{(a_1+a_2+..+a_{k-1}+a_{k+1}+...+a_n)}\ge \prod_{k=1}^n{\sqrt[n-1]{a_1a_2..a_{k-1}a_{k+1}...a_n}= (n-1)^na_1\cdot a_2...\cdot a_n \)