已知x>0,y>0,z>0,且x+y+z=1,求xyz的最大值? 已知XYZ是三个非负数，且满足3X+2Y+Z=5,2X+Y-...
\u5df2\u77e5\u5b9e\u6570x.y.z\u6ee1\u8db3x+y+z=0\u4e14xyz>0,\u8bb0T=1/x+1/y+1/z,\u5224\u65adT\u7684\u7b26\u53f7\u7531\u5df2\u77e5\u6613\u5f97
\u4e24\u8d1f\u4e00\u6b63
\u4e0d\u59a8\u8bbex>0>y,z
\u5219x=-y-z
T=(1/x)+(1/y)+(1/z)
=(1/-y-z)+(1/y)+(1/z)
=(1/y)-(1/y+z)+(1/z)
=z/[y(y+z)]+(1/z)<0
3X+2Y+Z=5,
2X+Y-3Z=1
\u89e3\u5f97x=7z-3, y=7-11z
\u2235XYZ\u662f\u4e09\u4e2a\u975e\u8d1f\u6570
\u2234x>=0\u53737z-3>=0
y>=0\u53737-11z>=0
z>=0
\u22343/7<=z<=7/11
K=3X+Y-7Z
=3(7z-3)+7-11z-7z
=3z-2
\u5f53z=3/7\u65f6\uff0cK\u6700\u5c0f\uff0c\u6700\u5c0f\u503c\u662f-5/7\uff0c\u5373b=-5/7
\u5f53z=7/11\u65f6\uff0cK\u6700\u5927\uff0c\u6700\u5927\u503c\u662f-1/11\uff0c\u5373a=-1/11
\u2234ab=5/77
呵呵，恩，基本不等式做，所以答案是1/27，当且仅当X=Y=z=1/3时取到最大值
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绛旓細x=1 浠ｅ叆 (2k+a)/3=2+(1-bk)/6 涓よ竟涔6 4k+2a=12+1-bk (4+b)k=13-2a 褰4+b=0涓13-2a=0鏃讹紝涓嶈k鍙栦綍鍊奸兘鎴愮珛 鎵浠=-4,a=13/2
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绛旓細缁撴灉涓猴細瀵逛簬浠绘剰瀹炴暟m锛屾柟绋嬮兘鏈変袱涓笉鐩哥瓑鐨勫疄鏁版牴 瑙ｉ杩囩▼濡備笅锛1銆佽嫢x=-1鏄柟绋嬬殑涓涓牴 鍙緱锛(-1)²-mX(-1)-2=0 瑙ｅ緱锛歮=1 浠ｅ叆鏂圭▼寰楋細x²-x-2=0 (x-2)(x+1)=0 瑙ｅ緱锛歺=2 鎴杧=-1 鎵浠ユ柟绋嬬殑鍙︿竴涓牴涓簒=2锛宮=1 2銆佲柍=m²-4x(-2)=m²...
