数学不等式证明,如图
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数学不等式证明,如图
![数学不等式证明,如图](/uploads/image/z/16604487-63-7.jpg?t=%E6%95%B0%E5%AD%A6%E4%B8%8D%E7%AD%89%E5%BC%8F%E8%AF%81%E6%98%8E%2C%E5%A6%82%E5%9B%BE%26nbsp%3B)
(1)
因为 x+y=1
所以xy≤1/4
(x+1/x)(y+1/y)=[(x^2+1)/x][(y^2+1)/y]
=(x^2+y^2+x^2*y^2+1)/xy
=x/y+y/x+xy+1/xy
=x/y+y/x+xy+(x+y)^2/xy
=2(x/y+y/x)+xy+2
≥6+xy=6+1/4=25/4
(2)
因为x+y=1
所以 xy≤1/4
2√(xy)≤x+y=1
x^2+y^2=(x+y)^2-2xy=1-2xy≥1/2
1/(xy)^2≥1/(1/4)^2 =16
(x+1/x)^2+(y+1/y)^2
=x^2+2+1/x^2 + y^2+2+1/y^2
=x^2+y^2 + (x^2+y^2)/(xy)^2 + 4
≥1/2+16/2+4 = 25/2
因为 x+y=1
所以xy≤1/4
(x+1/x)(y+1/y)=[(x^2+1)/x][(y^2+1)/y]
=(x^2+y^2+x^2*y^2+1)/xy
=x/y+y/x+xy+1/xy
=x/y+y/x+xy+(x+y)^2/xy
=2(x/y+y/x)+xy+2
≥6+xy=6+1/4=25/4
(2)
因为x+y=1
所以 xy≤1/4
2√(xy)≤x+y=1
x^2+y^2=(x+y)^2-2xy=1-2xy≥1/2
1/(xy)^2≥1/(1/4)^2 =16
(x+1/x)^2+(y+1/y)^2
=x^2+2+1/x^2 + y^2+2+1/y^2
=x^2+y^2 + (x^2+y^2)/(xy)^2 + 4
≥1/2+16/2+4 = 25/2