$\frac{a^{2}}{x}$ +$\frac{b^{2}}{y}$ + $\frac{c^{2}}{z}$ $\geq$ $\frac{\left ( a+b+c \right )^{2}}{x+y+z}$
Chứng minh BĐT
Started By TracyBloom, 20-12-2017 - 19:52
#1
Posted 20-12-2017 - 19:52
#2
Posted 20-12-2017 - 20:15
$\frac{a^{2}}{x}$ +$\frac{b^{2}}{y}$ + $\frac{c^{2}}{z}$ $\geq$ $\frac{\left ( a+b+c \right )^{2}}{x+y+z}$
Áp dụng Bunhia ta có:
(a^2/x+b^2/y+c^2/z)(x+y+z)>=(a+b+c)^2
=> đpcm
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