Giải bdt bằng pp toạ độ nhé!!!!!

H

huytrandinh

trong mặt phẳng tọa độ xét vector
[TEX]m=(\frac{1}{x^{2}},y^{2}),n=(\frac{1}{y^{2}},z^{2}),[/TEX]
[TEX]p=(\frac{1}{z^{2}},x^{2})[/TEX]
[TEX]x^{2}=\sqrt{a},y^{2}=\sqrt{b},z^{2}=\sqrt{c}[/TEX]
[TEX].|m|+|n|+|p|\geq |m+n+p|[/TEX]
[TEX]<=>VT\geq \sqrt{(x+y+z)^{2}+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}}[/TEX]
[TEX]\geq \sqrt{(x+y+z)^{2}+\frac{(1+1+1)^{4}}{(x+y+z)^{2}}}[/TEX]
[TEX]=\sqrt{(x+y+z)^{2}+\frac{81}{(x+y+z)^{2}}}\geq \sqrt{2.\sqrt{81}}[/TEX]
[TEX]=3\sqrt{2}[/TEX]
 
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