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	5	chỉnh		Sửa 17-04-16 03:35 PM ๖ۣۜJinღ๖ۣۜKaido 161 4
$\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ khi $x=\frac{1}{6};y=\frac{1}{3};z=\frac{1}{2}$chứng minh : $x,y>0$$(x+y)(\frac{a^2}{x}+\frac{b^2}{y})\geq (\sqrt{x}.\frac{a}{\sqrt{x}}+\sqrt{y}.\frac{b}{\sqrt{y}})^2=(a+b)^2$$\frac{a^2}{x}+\frac{b^2}{y}\geq \frac{(a+b)^2}{x+y}$-----> đây là schwarz ( gọi chung BCS) $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ khi $x=\frac{1}{6};y=\frac{1}{3};z=\frac{1}{2}$ $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ khi $x=\frac{1}{6};y=\frac{1}{3};z=\frac{1}{2}$chứng minh : $x,y>0$$(x+y)(\frac{a^2}{x}+\frac{b^2}{y})\geq (\sqrt{x}.\frac{a}{\sqrt{x}}+\sqrt{y}.\frac{b}{\sqrt{y}})^2=(a+b)^2$$\frac{a^2}{x}+\frac{b^2}{y}\geq \frac{(a+b)^2}{x+y}$-----> đây là schwarz ( gọi chung BCS)

	4	chỉnh		Sửa 17-04-16 10:48 AM ๖ۣۜJinღ๖ۣۜKaido 161 4
$\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ khi $x=\frac{1}{6};y=\frac{1}{3};z=\frac{1}{2}$ $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ đấu "=" tự tìm $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ khi $x=\frac{1}{6};y=\frac{1}{3};z=\frac{1}{2}$

	3	chỉnh		Sửa 17-04-16 10:46 AM ๖ۣۜJinღ๖ۣۜKaido 161 4
$\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ đấu "=" tự tìm $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$Vậy $min=36$ đấu "=" tự tìm $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$ ( cauchy schwarz)Vậy $min=36$ đấu "=" tự tìm

	2	chỉnh		Sửa 17-04-16 10:45 AM ๖ۣۜJinღ๖ۣۜKaido 161 4
$\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$Vậy $min=36$ đấu "=" tự tìm $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$Vậy $min=36$ $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$Vậy $min=36$ đấu "=" tự tìm

	1	tạo mới		Trả lời 17-04-16 10:45 AM ๖ۣۜJinღ๖ۣۜKaido 161 4
$\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq \frac{(1+2+3)^2}{x+y+z}=36$Vậy $min=36$

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