Cho ba số dương . Tìm giá trị nhỏ nhất của biểu thức sau : $P=\dfrac{a}{2a+5b+5c}+\dfrac{b}{2b+5c+5a}+\dfrac{c}{2b+5a+5b}$
$MinP=\dfrac{a}{2a+5b+5c}+\dfrac{b}{2b+5c+5a}+\dfrac{c}{2b+5a+5b}$
Started By thienminhdv, 15-09-2013 - 07:22
#1
Posted 15-09-2013 - 07:22
#2
Posted 15-09-2013 - 07:31
Theo bđt Bunhiacopxki ta có$= \frac{a^{2}}{2a^2+5ab+5ac}+\frac{b^2}{2b^2+5bc+5ba}+\frac{c^2}{2c^2+5ca+5bc}\geq \frac{(a+b+c)^2}{2(a^2+b^2+c^2)+10(ab+bc+ac)}\geq \frac{(a+b+c)^2}{2.(a+b+c)^2+6(ab+bc+ac)}\geq \frac{(a+b+c)^2}{2(a+b+c)^2+2(a+b+c)^2}= \frac{1}{4}$
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