Cho a,b,c>0 . CM : $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}+\frac{1}{2}(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\geq a+b+c$
$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}+\frac{1}{2}(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\geq
Started By deptrai9803, 06-01-2014 - 22:47
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Posted 06-01-2014 - 22:47
deptrai9803
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#2
Posted 06-01-2014 - 22:57
hoctrocuanewton
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Thiếu úy
áp dụng AM-GM ngược dấu ta có
$\sum \frac{a^{2}}{a+b}+\frac{1}{2}\sum \sqrt{ab}=\sum a-\sum \frac{ab}{a+b}+\frac{1}{2}\sqrt{ab}\geq \sum a-\sum \frac{ab}{2\sqrt{ab}}+\frac{1}{2}\sqrt{ab}=\sum a$
$\Rightarrow Q.E.D$
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