若x＞0，y＞0且x＋y＝1，则1／y＋2／x的最小值为 已知x>0,y>0,且1/x+2/y=2,求x+y最小值
\u5df2\u77e5x\uff1e0\uff0cy\uff1e0\uff0cx+y=1\uff0c\u52191/x+1/2y\u7684\u6700\u5c0f\u503c\u4e3a\uff1f(1/x+1/2y)=(1/x+1/2y)(x+y)
=1+y/x+x/2y+1/2
=3/2+y/x+x/2y
\u22673/2+2\u221a(1/2)
=3/2+\u221a2
\u5f53\u4e14\u4ec5\u5f53y/x=x/2y\u65f6\uff0c\u7b49\u53f7\u6210\u7acb\u3002
\u795d\u4f60\u5f00\u5fc3\uff01\u5e0c\u671b\u80fd\u5e2e\u5230\u4f60\uff0c\u5982\u679c\u4e0d\u61c2\uff0c\u8bf7\u8ffd\u95ee\uff0c\u795d\u5b66\u4e60\u8fdb\u6b65\uff01O(\u2229_\u2229)O
\u56e0\u4e3a1/x+2/y=2
\u6240\u4ee5x=y/(2y-2)
x+y
=y/(2y-2)+y
=0.5y/(y-1)+(y-1)+1
=0.5[(y-1)+1)]/(y-1)+(y-1)+1
=0.5/(y-1)+(y-1)+1.5
\u2265sqrt(2)+1.5
\u6b64\u65f6y=1+sqrt(2)/2\u3002
x+y\u6700\u5c0f\u503c\u4e3asqrt(2)+1.5
则1／y＋2／x
=(1/y+2/x)(x+y)
=x/y+1+2+2y/x
=3+x/y+2y/x≥3+2√(x/y)×(2y/x)=3+2√2;
所以最小值=3+2√2;
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