Daca a,b,c,d reale sunt astfel incat
\( a+b\sqrt{2}+c\sqrt{3}+d\sqrt{4}\ge \sqrt{10(a^2+b^2+c^2+d^2)} \)
aratati ca \( a^2+d^2=b^2+c^2 \)
Concursul "Cezar Ivanescu" 2008
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Marius Mainea
- Gauss
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- Mateescu Constantin
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Aplicam inegalitatea C.B.S : \( (a+b\sqrt 2+c\sqrt 3+d\sqrt 4)^2\le (a^2+b^2+c^2+d^2)(1+2+3+4) \)
\( \Longleftrightarrow\ |a+b\sqrt 2+c\sqrt 3+d\sqrt 4|\le \sqrt{10(a^2+b^2+c^2+d^2)} \) si conform enuntului suntem in cazul de egalitate .
Asadar \( \frac a1=\frac{b}{sqrt 2}=\frac c{\sqrt 3}=\frac{d}{\sqrt 4} \) , iar egalitatea de demonstrat se verifica usor .
\( \Longleftrightarrow\ |a+b\sqrt 2+c\sqrt 3+d\sqrt 4|\le \sqrt{10(a^2+b^2+c^2+d^2)} \) si conform enuntului suntem in cazul de egalitate .
Asadar \( \frac a1=\frac{b}{sqrt 2}=\frac c{\sqrt 3}=\frac{d}{\sqrt 4} \) , iar egalitatea de demonstrat se verifica usor .
Last edited by Mateescu Constantin on Mon Dec 28, 2009 5:38 pm, edited 1 time in total.