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Inegalitate (easy!)
Posted: Fri Jan 02, 2009 6:38 pm
by maxim bogdan
Fie \( a,b,c\in\mathbb{R} \) astfel incat \( \sum_{cyc}\frac{1}{1+a^2}=2. \) Demonstrati ca este satisfacuta inegalitatea:
\( abc(a+b+c-abc)\leq\frac{5}{8}. \)
Posted: Sat Jan 03, 2009 2:28 am
by Marius Mainea
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 \( LHS\le4\sum{x(yx+yz+zx+zy)}=24xyz+4\sum{xy(x+y)}=18xyz+6xyz+4\sum{xy(x+y)}\le18xyz+\sum{xy(x+y)}+4\sum{xy(x+y)}=RHS \)