cho xy+yz+xz=671
cm$\frac{y}{y^2-xz+2013}+\frac{z}{z^2-xy+2013}+\frac{x}{x^2-yz+2013}\geq \frac{1}{x+y+z}$
Edited by viendanho98, 13-07-2013 - 14:50.
cm $\frac{y}{y^2-xz+2013}+\frac{z}{z^2-xy+2013}+\frac{x}{x^2-yz+2013} \geq \frac{1}{x+y+z}$
Started By viendanho98, 13-07-2013 - 14:47
#2
Posted 13-07-2013 - 15:54
Best Friend
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cho xy+yz+xz=671
cm$P=\frac{y}{y^2-xz+2013}+\frac{z}{z^2-xy+2013}+\frac{x}{x^2-yz+2013}\geq \frac{1}{x+y+z}$
Áp dụng BĐT Am-Gm
Ta có : $P=\frac{y}{y^2-xz+2013}+\frac{z}{z^2-xy+2013}+\frac{x}{x^2-yz+2013}=\frac{y^{2}}{y^{3}-xyz+2013y}+\frac{z^{2}}{z^{3}-xyz+2013z}+\frac{x^{2}}{x^{3}-xyz+2013x}\geq \frac{(x+y+z)^{2}}{\sum x^{3}-3xyz+2013(y+x+z)}=\frac{(x+y+z)^{2}}{(x+y+z)(\sum x^{2}-\sum xy)+(3\sum xy)(x+y+z)}=\frac{(x+y+z)^{2}}{(x+y+z)(\sum x^{2}+2\sum xy)}=\frac{x+y+z}{(\sum x)^{2}}=\frac{1}{x+y+z}$
Dấu = xảy ra khi ..............
Edited by Best Friend, 13-07-2013 - 16:00.
Best Friend 
#3
Posted 14-07-2013 - 09:35
quangtq1998
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Ta có : $VT \ge \frac{(x+y+z)^2}{x^3+y^3+z^3-3xyz+2013(x+y+z) } $
$= \frac{(x+y+z)^2}{x^3+y^3+z^3-3xyz+3(xy+yz+xz) (x+y+z) }$
$= \frac{x+y+z)^2}{x^3+y^3+z^3+3(x+y)(y+z)(x+z) }$
$= \frac{x+y+z)^2}{(x+y+z)^3} = \frac{1}{x+y+z} $
Dấu "=" xảy ra khi .....
$= \frac{(x+y+z)^2}{x^3+y^3+z^3-3xyz+3(xy+yz+xz) (x+y+z) }$
$= \frac{x+y+z)^2}{x^3+y^3+z^3+3(x+y)(y+z)(x+z) }$
$= \frac{x+y+z)^2}{(x+y+z)^3} = \frac{1}{x+y+z} $
Dấu "=" xảy ra khi .....
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