不等式 3(學科能力競賽)

令 a,b,c 是正數 且a=x^3 b=y^3 c=z^3, 所以x,y,z 是正數
又abc=8 , so , xyz=2
2+a+b=xyz+x^3+y^3
因為(x-y)^2大於等於0 所以
x^2-xy+y^2大於等於xy
(x+y)(x^2-xy+y^2)大於等於(x+y)xy
x^3+y^3大於等於(x^2)y+x(y^2)
2+a+b=xyz+x^3+y^3大於等於xyz+(x^2)y+x(y^2)=xy(x+y+z)
1/(2+a+b) =(1/2)[2/(2+a+b)]=(1/2)[(xyz)/(2+a+b)]
=(1/2)[(xyz)/(xyz+x^3+y^3)]小於等於(1/2)[(xyz)/(xy(x+y+z))]
=(1/2)[z/(x+y+z))]

同理1/(2+b+c) 小於等於(1/2)[x/(x+y+z))]
1/(2+c+a) 小於等於(1/2)[y/(x+y+z))]
所以[1/(2+a+b) ] +[1/(2+b+c)] +[ 1/(2+c+a)]小於等於(1/2)[z/(x+y+z))]
+(1/2)[x/(x+y+z))]+(1/2)[y/(x+y+z))] =(1/2)

Rearrangement inequality: (x^2)(x)+(y^2)(y) >= (x^2)(y)+(y^2)(x)