当$x,yge0,x+y=2$时求下面式子的最小值:
1)$x+sqrt{x^2-2x+y^2+1}$
2)$dfrac{1}{5}x+sqrt{x^2-2x+y^2+1}$
解:1)$P(x,y)$为直线$x+y=2$上一点,点$H$为$P$到$y$轴的投影点,
设$A(1,0)$则$A$关于$x+y=2$的对称点$A'(2,1)$
故$x+sqrt{x^2-2x+y^2+1}=|PH|+|PA|= |PH|+|PA'|ge2$
2)$dfrac{1}{5}x+sqrt{x^2-2x+y^2+1}$
$=dfrac{1}{5}x+sqrt{(x^2-2x+y^2+1)(cos^2 heta+sin^2 heta)}$
$ge(dfrac{1}{5}+cos heta)x+ysin heta-cos heta$
令$cos heta=dfrac{3}{5},sin heta=dfrac{4}{5}$则最小值为1