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Hong Kong 1997

Started By euler, 01-09-2005 - 08:03

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#1
euler

Posted 01-09-2005 - 08:03

Thượng sĩ

Thành viên
275 posts

Cho http://dientuvietnam.net/cgi-bin/mimetex.cgi?\large\dfrac{xyz(x+y+z+\sqrt{x^2+y^2+z^2})}{(x^2+y^2+z^2)(xy+yz+zx)}\le\dfrac{3+\sqrt{3}}{9}

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địa chỉ nào cũng được!

#2
tk14nkt

Posted 01-09-2005 - 09:52

Đồi gió hú

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358 posts

Đặt vế trái của bđt đã cho là P.
Dễ có:
Từ các bđt quen thuộc:
$x+y+z \leq \sqrt{3(x^2+y^2+z^2)}$
$x^2+y^2+z^2 \geq 3\sqrt[3]{(xyz)^2}$
$xy+yz+zx\geq 3\sqrt[3]{(xyz)^2}$ .
Suy ra:
$\dfrac{1}{3} \geq \dfrac{xyz(x+y+z)}{(x^2+y^2+z^2)(xy+yz+zx)}$
$\dfrac{\sqrt{3}}{9}\geq \dfrac{xyz\sqrt{x^2+y^2+z^2}}{(x^2+y^2+z^2)(xy+yz+zx)}$
Cộng theo vế ta có đfcm.
Đẳng thức có...

Trying not to break

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Hong Kong 1997

#1 euler Posted 01-09-2005 - 08:03

#2 tk14nkt Posted 01-09-2005 - 09:52

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#1
euler

Posted 01-09-2005 - 08:03

#2
tk14nkt

Posted 01-09-2005 - 09:52