不等式 比大小x^2+y^2=1和2(x+y-1) x^2+5x+6和2x^2+5x+9当x>1时,x^3和x^2-x+1已知a>b>0,c>d>0求证根号a/d>根号b/cx^2+y^2=1改x^2+y^2+1

Question

不等式 比大小
x^2+y^2=1和2(x+y-1)
x^2+5x+6和2x^2+5x+9
当x>1时,x^3和x^2-x+1
已知a>b>0,c>d>0求证根号a/d>根号b/c
x^2+y^2=1
改
x^2+y^2+1

Answer 1

x^2+y^2+1-2(x+y-1)
=(x^2+2x+1)+(y^2+2y+1)+1
=(x+1)^2+(y+1)^2+1>0
所以,x^2+y^2+1>2(x+y-1)
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x^2+5x+6-(2x^2+5x+9 )
=-x^2-3=-(x^2+3)1)
所以,x^3>(x^2-x+1)
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a>b>0,√a>√b
c>d>0,√c>√d
√ac>√bd
√a/d>√b/c