证明√X2+Y2+√(X-1)2+Y2+√X2+(Y-1)2+√(X-1)2+(Y-1)2>=2√2并求=成立时X与Y的值_数学解答

证明√X2+Y2+√(X-1)2+Y2+√X2+(Y-1)2+√(X-1)2+(Y-1)2>=2√2并求=成立时X与Y的值

人气：162 ℃　时间：2020-06-12 01:46:27

解答

√X2+Y2+√(X-1)2+Y2+√X2+(Y-1)2+√(X-1)2+(Y-1)2>=2√2
在\x\=\y\,\x-1\=y,\y-1\=\x\,\x-1\=\y-1\时值最小
解得x=1/2,y=1/2
代入原式:
√X2+Y2+√(X-1)2+Y2+√X2+(Y-1)2+√(X-1)2+(Y-1)2
=4根号(1/2)
=2根号2
所以x=1/2,y=1/2