Cho $x,y,z>0$ và $2x+4y+7z=2xyz$. Tìm GTNN $P=x+y+z$
#1
Posted 12-03-2013 - 23:41
#2
Posted 13-03-2013 - 11:40
$x+y+z=3.\frac{x}{3}+\frac{5}{2}.\frac{2y}{5}+2.\frac{c}{2}\geq (3+2,5+2).\left [ (\frac{x}{3})^3.(\frac{y}{2,5})^{2,5}.(\frac{z}{2}^2) \right ]^{\frac{1}{3+2,5+2}}$Cho $x,y,z>0$ và $2x+4y+7z=2xyz$. Tìm GTNN $P=x+y+z$
$2x+4y+7z=2.3.\frac{x}{3}+4.2,5.\frac{y}{2,5}+7.2.\frac{z}{2}\geq (2.3+4.2,5+7.2)\left [ (\frac{x}{3})^{2.3}.(\frac{y}{2,5})^{4.2,5}.(\frac{z}{2})^{7.2} \right ]^{\frac{1}{2.3+4.2,5+7.2}}=30.\left [ (\frac{x}{3})^6.(\frac{y}{2,5})^{10}.(\frac{z}{2})^{14} \right ]^{\frac{1}{30}}$
$\Rightarrow (x+y+z)^2.(2x+4y+7z)\geq (3+2,5+2)^2.(2.3+4.2,5+7.2).(\frac{x}{3})^{\frac{2.2}{3+2,5+2}+\frac{2.2}{2.3+4.2,5+7.2}}.(\frac{y}{2,5})^{\frac{2.2,5}{3+2,5+2}+\frac{4.2,5}{2.3+4.2,5+7.2}}.(\frac{z}{2})^{\frac{2.2}{3+2,5+2}+\frac{7.2}{2.3+4.2,5+7.2}}=(3+2,5+2)^2.(2.3+4.2,5+7.2).(\frac{x}{3}).(\frac{y}{2,5}).(\frac{z}{2})=2.(\frac{15}{2})^2.xyz$
$\Rightarrow (x+y+z)^2\geq (\frac{15}{2})^2$
$\Rightarrow x+y+z \geq 7,5$
Đẳng thức xảy ra khi và chỉ khi x=3, y=2,5, z=2
Edited by vutuanhien, 13-03-2013 - 11:45.
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#3
Posted 16-03-2013 - 13:24
Mình đã hiểu cách làm nhưng mình còn thắc mắc cơ sở tại sao bạn lại tách thành $\frac{x}{3},\frac{y}{2,5},\frac{z}{2}$.$x+y+z=3.\frac{x}{3}+\frac{5}{2}.\frac{2y}{5}+2.\frac{c}{2}\geq (3+2,5+2).\left [ (\frac{x}{3})^3.(\frac{y}{2,5})^{2,5}.(\frac{z}{2}^2) \right ]^{\frac{1}{3+2,5+2}}$
$2x+4y+7z=2.3.\frac{x}{3}+4.2,5.\frac{y}{2,5}+7.2.\frac{z}{2}\geq (2.3+4.2,5+7.2)\left [ (\frac{x}{3})^{2.3}.(\frac{y}{2,5})^{4.2,5}.(\frac{z}{2})^{7.2} \right ]^{\frac{1}{2.3+4.2,5+7.2}}=30.\left [ (\frac{x}{3})^6.(\frac{y}{2,5})^{10}.(\frac{z}{2})^{14} \right ]^{\frac{1}{30}}$
$\Rightarrow (x+y+z)^2.(2x+4y+7z)\geq (3+2,5+2)^2.(2.3+4.2,5+7.2).(\frac{x}{3})^{\frac{2.2}{3+2,5+2}+\frac{2.2}{2.3+4.2,5+7.2}}.(\frac{y}{2,5})^{\frac{2.2,5}{3+2,5+2}+\frac{4.2,5}{2.3+4.2,5+7.2}}.(\frac{z}{2})^{\frac{2.2}{3+2,5+2}+\frac{7.2}{2.3+4.2,5+7.2}}=(3+2,5+2)^2.(2.3+4.2,5+7.2).(\frac{x}{3}).(\frac{y}{2,5}).(\frac{z}{2})=2.(\frac{15}{2})^2.xyz$
$\Rightarrow (x+y+z)^2\geq (\frac{15}{2})^2$
$\Rightarrow x+y+z \geq 7,5$
Đẳng thức xảy ra khi và chỉ khi x=3, y=2,5, z=2
Bạn có thể giải thích rõ cho mình để mình có thể áp dụng làm những bài tương tự thế này được không?
Cảm ơn trước!
Edited by jb7185, 16-03-2013 - 13:25.
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