Fie \( a,b,c \) numere reale strict pozitive, cu \( a+b+c=3. \) Sa se demonstreze inegalitatea:
\( \frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\geq 3. \)
INEGALITATE BARAJ JUNIORI 2009
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maxim bogdan
- Thales
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INEGALITATE BARAJ JUNIORI 2009
Feuerbach
-
maxim bogdan
- Thales
- Posts: 106
- Joined: Tue Aug 19, 2008 1:56 pm
- Location: Botosani
Solutia mea
Avem:
\( \displaystyle\sum_{cyc}\frac{a+3}{3a+bc}=\displaystyle\sum_{cyc}\frac{(a+b)+(a+c)}{(a+b)(a+c)} \)
Introducem notatiile: \( x=a+b;y=b+c;z=c+a. \) Acum vom avea: \( x+y+z=6. \) Inegalitatea se rescrie:
\( \displaystyle\sum_{cyc}\frac{x+y}{xy}=2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 2\cdot\frac{9}{x+y+z}=2\cdot\frac{9}{6}=3. \)
Mai sus am aplicat inegalitatea Cauchy-Schwarz.
\( \displaystyle\sum_{cyc}\frac{a+3}{3a+bc}=\displaystyle\sum_{cyc}\frac{(a+b)+(a+c)}{(a+b)(a+c)} \)
Introducem notatiile: \( x=a+b;y=b+c;z=c+a. \) Acum vom avea: \( x+y+z=6. \) Inegalitatea se rescrie:
\( \displaystyle\sum_{cyc}\frac{x+y}{xy}=2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 2\cdot\frac{9}{x+y+z}=2\cdot\frac{9}{6}=3. \)
Mai sus am aplicat inegalitatea Cauchy-Schwarz.
Feuerbach