Bài toán khó đây đề ra rất ngắn

H

hunterking

theo BDT bunhia ta có:
P=[TEX]\sqrt{x} +2\sqrt{y}[/TEX]\leq [TEX] \sqrt ({1^2+2^2})({x+y})[/TEX]
\Leftrightarrow 10\leq [TEX] \sqrt ({1^2+2^2})({x+y})[/TEX]
\Leftrightarrow [TEX]\frac{10}{\sqrt {5}}[/TEX]\leq [TEX]\sqrt {x+y}[/TEX]
\Leftrightarrow x+y\geq 20 (bình phương 2 vế)\Rightarrow dpcm
dấu = có \Leftrightarrow y=4x
\Leftrightarrow x=4
\Leftrightarrow y=16
 
T

tuatprohd

Thanks very much......................................................................................................................................:D:D:D:D
 
You must log in or register to reply here.
Top Bottom