Fie \( a,b,c\in R \) cu \( |a|>|b|\ge|c| \). Demonstrati ca:
\( \sqrt{\frac{a^2+b^2}{a^2-b^2}}+\sqrt{\frac{a^2+c^2}{a^2-c^2}}\ge 2(1+\frac{|bc|}{a^2}) \)
In ce caz are loc egalitatea?
ETAPA LOCALA 2008, DOLJ
inegalitate interesanta
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BogdanCNFB
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BogdanCNFB
- Thales
- Posts: 121
- Joined: Wed May 07, 2008 4:29 pm
- Location: Craiova
Din \( Mg\ge Mh \) avem \( \sqrt{1\cdot\frac{a^2+b^2}{a^2-b^2}}\ge \frac{2}{1+\frac{a^2-b^2}{a^2+b^2}}=\frac{a^2+b^2}{a^2}=1+\frac{b^2}{a^2} \)
Analog, \( \sqrt{\frac{a^2+c^2}{a^2-c^2}}\ge 1+\frac{c^2}{a^2} \)
Prin insumare, rezulta \( \sqrt{\frac{a^2+b^2}{a^2-b^2}}+\sqrt{\frac{a^2+c^2}{a^2-c^2}}\ge 2+\frac{b^2+c^2}{a^2}\ge 2+\frac{2|bc|}{a^2}=2(1+\frac{|bc|}{a^2}) \)
Evident, egalitatea are loc pentru \( b=c=0 \)
Analog, \( \sqrt{\frac{a^2+c^2}{a^2-c^2}}\ge 1+\frac{c^2}{a^2} \)
Prin insumare, rezulta \( \sqrt{\frac{a^2+b^2}{a^2-b^2}}+\sqrt{\frac{a^2+c^2}{a^2-c^2}}\ge 2+\frac{b^2+c^2}{a^2}\ge 2+\frac{2|bc|}{a^2}=2(1+\frac{|bc|}{a^2}) \)
Evident, egalitatea are loc pentru \( b=c=0 \)