Fie \( a,b,c\in\mathbb{R} \) astfel incat \( \sum_{cyc}\frac{1}{1+a^2}=2. \) Demonstrati ca este satisfacuta inegalitatea:
\( abc(a+b+c-abc)\leq\frac{5}{8}. \)
Inegalitate (easy!)
Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea
-
maxim bogdan
- Thales
- Posts: 106
- Joined: Tue Aug 19, 2008 1:56 pm
- Location: Botosani
Inegalitate (easy!)
Feuerbach
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Deconditionam cu \( a=\sqrt{\frac{x}{y+z}} \) si analoagele si inegalitatea devine
\( \sqrt{\frac{xyz}{(x+y)(y+z)(z+x)}}\cdot\sum{\sqrt{\frac{x}{y+x}}}\le\frac{5}{8}+\frac{xyz}{(x+y)(y+z)(z+x)} \)
sau
\( 8\sqrt{xyz}\sum{\sqrt{x(x+y)(x+z)}}\le 5(x+y)(y+z)(z+x)+8xyz \)
sau
\( 8\sum{{x\sqrt{[y(x+z)][z(x+y)]}}\le18xyz+5\sum{xy(x+y)} \)
Insa din inegalitatea AM-GM \( LHS\le4\sum{x(yx+yz+zx+zy)}=24xyz+4\sum{xy(x+y)}=18xyz+6xyz+4\sum{xy(x+y)}\le18xyz+\sum{xy(x+y)}+4\sum{xy(x+y)}=RHS \)
\( \sqrt{\frac{xyz}{(x+y)(y+z)(z+x)}}\cdot\sum{\sqrt{\frac{x}{y+x}}}\le\frac{5}{8}+\frac{xyz}{(x+y)(y+z)(z+x)} \)
sau
\( 8\sqrt{xyz}\sum{\sqrt{x(x+y)(x+z)}}\le 5(x+y)(y+z)(z+x)+8xyz \)
sau
\( 8\sum{{x\sqrt{[y(x+z)][z(x+y)]}}\le18xyz+5\sum{xy(x+y)} \)
Insa din inegalitatea AM-GM \( LHS\le4\sum{x(yx+yz+zx+zy)}=24xyz+4\sum{xy(x+y)}=18xyz+6xyz+4\sum{xy(x+y)}\le18xyz+\sum{xy(x+y)}+4\sum{xy(x+y)}=RHS \)