Fie a,b,c>0 cu \( \sum{\frac{a}{b+c+1}}=1 \). Demonstrati ca
\( \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\ge 1 \)
Inegalitate rationala conditionata
Moderators: Bogdan Posa, Laurian Filip
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
-
Claudiu Mindrila
- Fermat
- Posts: 520
- Joined: Mon Oct 01, 2007 2:25 pm
- Location: Targoviste
- Contact:
Solutie. Fara a leza generalitatea putem presupune ca \( a= \min (a,b,c) \). Atunci \( a\le b\le c \) si \( \frac{1}{b+c+1}\le\frac{1}{c+a+1}\le\frac{1}{a+b+1} \). Scriem inegalitatea rearanjamentelor de doua ori si obtinem:
\( \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}=1 (1) \)
si
\( \frac{b}{a+b+1}+\frac{c}{b+c+1}+\frac{a}{c+a+1}\le\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}=1 (2) \).
Din \( (1) \) si \( (2) \) avem:
\( \frac{a+b}{a+b+1}+\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\le2\Leftrightarrow3-\left(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\right)\le2\Leftrightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\ge1 . \qed \)
\( \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}=1 (1) \)
si
\( \frac{b}{a+b+1}+\frac{c}{b+c+1}+\frac{a}{c+a+1}\le\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}=1 (2) \).
Din \( (1) \) si \( (2) \) avem:
\( \frac{a+b}{a+b+1}+\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\le2\Leftrightarrow3-\left(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\right)\le2\Leftrightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\ge1 . \qed \)
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste