Din nou o inegalitate
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Claudiu Mindrila
- Fermat
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Din nou o inegalitate
Fie \( a,b,c \) lungimile laturilor unui triunghi. Aratati ca: \( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} < \frac{5}{a+b+c} \)
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
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Marius Mainea
- Gauss
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- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
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
\( \sum {\frac{x}{1+x}}=\sum {\frac{x^2}{x+x^2}}\geq\frac{(x+y+z)^2}{x+y+z+x^2+y^2+z^2}=\frac{1}{1+x^2+y^2+z^2}>\frac{1}{1+x+y+z}=\frac{1}{2} \)
\( \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}<\frac{5}{2} \)
sau \( \frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}>\frac{1}{2} \),
care este adevarata conform CBS
\( \sum {\frac{x}{1+x}}=\sum {\frac{x^2}{x+x^2}}\geq\frac{(x+y+z)^2}{x+y+z+x^2+y^2+z^2}=\frac{1}{1+x^2+y^2+z^2}>\frac{1}{1+x+y+z}=\frac{1}{2} \)