求外切于单位圆的直角三角形的周长的最小值

解：设三角形ABC 角C是直角
O为圆心 切点分别是E F G
E在AB上 F在BC上 G在AC上
所以CF=CG=1
设FB=x AG=y
所以EB=FB=x AE=AG=y
所以周长为2x+2y+2
因为是直角三角形 所以满足(x+1)^2+(y+1）^2=(x+y)^2
所以2x+2y+2=2xy
所以x+y+1=xy
y=(x+1)/(x-1)
周长=2x+2y+2
=2*(x+y+1)
=2*(x+1+(x+1)/(x-1))
=2*((x^2-1+x+1)/(x-1))
=2*(x^2+x)/(x-1)
=2*(x^2-x+2x-2+2)/(x-1)
=2*(x+2+2/(x-1))
=2*(x-1+2/(x-1)+3)
>=2*(2*√((x-1)*2/(x-1))+3) 当x-1=2/(x-1)时等号成立
=2*(2√2+3)
=4√2+6
此时x=y=√2+1