Question

已知非负实数x,yz满足x+y+z=1,则t=2xy+yz+2zx的最大值为多少？

Answer 1

y+z=1-x≥2√yz
yz≤(1-x)^2/4≤t=2xy+yz+2zx=2x(y+z)+yz≤2x(1-x)+(1-x)^2/4
4t≤8x-8x^2+x^2-2x+1=6x-7x^2+1
7x²-6x+4t-1≤0
△≥0
36-28*(4t-1)≥0
64-112t≥0
t≤4/7

http://wenku.baidu.com/view/d7de2f94c1c708a1284a44fe.html
这上面也有答案

Answer 2

y+z=1-x>=2√yz
yz<=(1-x)^2/4

t=2xy+yz+2zx
=2x(y+z)+yz
<=2x(1-x)+(1-x)^2/4
4t<=8x-8x^2+x^2-2x+1
=6x-7x^2+1
7x^2-6x+4t-1<=0
△>=0
36-28*(4t-1)>=0
74-112t>=0
t<=37/56

Answer 3

七分之四