Cho $a,b,c>0$ và $a+b+c=1$
Chứng minh rằng $\sum \frac{\sqrt{a^{2}+2ab}}{\sqrt{b^{2}+2c^{2}}}\geq \frac{1}{\sum a^{2}}$
$\sum \frac{\sqrt{a^{2}+2ab}}{\sqrt{b^{2}+2c^{2}}}\geq \frac{1}{\sum a^{2}}$
Started By duaconcuachua98, 03-08-2014 - 20:19
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Posted 04-08-2014 - 16:22
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Thiếu tá
Cho $a,b,c>0$ và $a+b+c=1$
Chứng minh rằng $\sum \frac{\sqrt{a^{2}+2ab}}{\sqrt{b^{2}+2c^{2}}}\geq \frac{1}{\sum a^{2}}$
Theo AM-GM có $\sum \frac{\sqrt{a^2+2ab}}{\sqrt{b^2+2c^2}}=\sum \frac{a^2+2ab}{\sqrt{(a^2+2ab)(b^2+2c^2)}}\geq\sum \frac{a^2+2ab}{\frac{a^2+2ab+b^2+2c^2}{2}}\geq \sum \frac{a^2+2ab}{\frac{2(a^2+b^2+c^2)}{2}}=\sum \frac{a^2+2ab}{a^2+b^2+c^2}=\frac{(\sum a)^2}{\sum a^2}=\frac{1}{\sum a^2}$
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