Cho x,y dương thỏa mãn x+y=1 . CMR : $8(x^{4}+y^{4})+\frac{1}{xy}\geq 5$
CMR : $8(x^{4}+y^{4})+\frac{1}{xy}\geq 5$
Started By congdaoduy9a, 24-04-2015 - 22:55
#1
Posted 24-04-2015 - 22:55
#2
Posted 24-04-2015 - 23:02
$8(x^4+y^4)+\frac{1}{xy}\geq (x+y)^4+\frac{4}{(x+y)^2}=5$
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FAN THẦY THÔNG,ANH CẨN,THẦY VINH
#3
Posted 24-04-2015 - 23:07
áp dụng BĐT AM-GM ta có:
$8(x^{4}+y^{4})+\frac{1}{xy}\geq 16x^{2}y^{2}+\frac{1}{xy}=16x^{2}y^{2}+\frac{1}{4xy}+\frac{1}{4xy}+\frac{1}{2xy}\geq 3\sqrt[3]{16x^{2}y^{2}.\frac{1}{4xy}.\frac{1}{4xy}}+\frac{2}{(x+y)^{2}}=3+2=5$
Dấu bằng xảy ra khi và chỉ khi x=y=1/2
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#4
Posted 24-04-2015 - 23:08
$8(x^{4}+y^{4})\geqslant 8\frac{(x^{2}+y^{2})^{2}}{2}\geqslant 8\frac{(x+y)^{4}}{8}=1$
1$\geqslant 2\sqrt{xy}=>xy\leqslant \frac{1}{4}=>\frac{1}{xy}\geqslant 4$
=>$8(x^{4}+y^{4})+\frac{1}{xy}\geqslant 5$
Dấu "=" xảy ra khi x=y=0,5
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