Cho $x,y,z>0$ và $x^2+y^2+z^2=1$
Tìm Min $$\frac{x}{(1-x^4)2}+\frac{y}{(1-y^4)^2}+\frac{z}{(1-z^4)^2}$$
Min $$\sum \frac{x}{(1-x^4)^2}$$
Started By gianglt1701, 30-04-2016 - 01:06
#2
Posted 30-04-2016 - 07:55
quangtq1998
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Cho $x,y,z>0$ và $x^2+y^2+z^2=1$
Tìm Min $$\frac{x}{(1-x^4)2}+\frac{y}{(1-y^4)^2}+\frac{z}{(1-z^4)^2}$$
$\frac{t}{(1-t^4)^2} \ge \frac{81t^2\sqrt{3}}{64} $
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