若实数x>0,y>0,满足x+2y=1,则xy最大值是？？用基本不等式解。。。急 若实数xy满足xy>0,则x/(x+y)+2y/(x+2y)...
\u82e5x>0,y>0,\u4e14x+2y=10\uff0c\u5219xy\u7684\u6700\u5927\u503c\u662f
\u8bf7\u91c7\u7eb3
\u89e3\uff1a\u53ef\u4ee4x+y=s\uff0cx+2y=t\uff0c
\u7531xy\uff1e0\uff0c\u53ef\u5f97x\uff0cy\u540c\u53f7\uff0cs\uff0ct\u540c\u53f7\uff0e
\u5373\u6709x=2s-t\uff0cy=t-s\uff0c
\u5219x/(x+y)+2y/(x+2y)=(2s-t)/s+(2t-2s)/t
=4-\uff08t/s+2s/t\uff09\u22644-2\u221a2
\u5f53\u4e14\u4ec5\u5f53t^2=2s^2\uff0c\u53d6\u5f97\u7b49\u53f7\uff0c
\u5373\u6709\u6240\u6c42\u6700\u5927\u503c\u4e3a4-\u221a2\uff0e
解xy=1/2×x×2y
≤1/2[(x+2y)/2]^2
=1/2×（1/2)^2
=1/8
当且仅当x=2y时，等号成立
即当且仅当x=1/2，y=1/4时，等号成立
故xy的最大值为1/8.
利用均值不等式
xy = (1/2)(x*2y) ≤ (1/2)[(x+2y)/2]^2 = ……
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