Cho a, b, c dương. CMR: $\sum \frac{a^{2}}{b}\geqslant \sum a+\frac{4(a-c)^{2}}{a+b+c}$
Edited by buitudong1998, 18-03-2014 - 07:40.
CMR: $\sum \frac{a^{2}}{b}\geqslant \sum a+\frac{4(a-c)^{2}}{a+b+c}$
Started By buitudong1998, 18-03-2014 - 07:39
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Posted 18-03-2014 - 12:46
kfcchicken98
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$\sum \frac{a^{2}}{b}-2a+b=\sum \frac{(a-b)^{2}}{b}\geq \frac{(\left | a-b \right |+\left | b-c \right |+\left | c-a \right |)^{2}}{a+b+c}\geq \frac{(\left | a-c \right |+\left | c-a \right |)^{2}}{a+b+c}=\frac{4\left | a-c \right |^{2}}{a+b+c}$
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