Question

cho a>b> 0 so sánh x, y với:

x=\(\frac{1+a}{1+a+a^2}\) và y=\(\frac{1+b}{1+b+b^2}\)

Answer 1

\(x-y=A=\frac{1+a}{1+a+a^2}-\frac{1+b}{1+b+b^2}=\frac{\left(1+a\right)\left(1+b+b^2\right)-\left(1+b\right)\left(1+a+a^2\right)}{\left(1+a+a^2\right)\left(1+b+b^2\right)}\)

\(A=\frac{\left(1+b+b^2+a+ab+ab^2\right)-\left(1+a+a^2+b+ab+a^2b\right)}{\left(1+a+a^2\right)\left(1+b+b^2\right)}=\frac{ab^2-a^2b}{\left(1+a+a^2\right)\left(1+b+b^2\right)}\)

\(A=\frac{ab\left(b-a\right)}{\left(1+a+a^2\right)\left(1+b+b^2\right)}< 0\) do a>b>0; mẫu>0

Vậy \(x-y< 0\Rightarrow x< y\)