Din nou o inegalitate
Posted: Sat Jul 19, 2008 5:18 pm
Fie \( a,b,c \) lungimile laturilor unui triunghi. Aratati ca: \( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} < \frac{5}{a+b+c} \)
Notam a=x+y, b=y+z ,c=z+x ,x>0, y>0 ,z>0.Prin conditionare, putem presupune ca x+y+z=1 si inegalitatea devine :
\( \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}<\frac{5}{2} \)
sau \( \frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}>\frac{1}{2} \),
care este adevarata conform CBS
\( \sum {\frac{x}{1+x}}=\sum {\frac{x^2}{x+x^2}}\geq\frac{(x+y+z)^2}{x+y+z+x^2+y^2+z^2}=\frac{1}{1+x^2+y^2+z^2}>\frac{1}{1+x+y+z}=\frac{1}{2} \)