Question

Cho a>b>0. Biết x=(a+1)/(a^2+a+1); y=(b+1)/(b^2+b+1). So sánh a và b

Answer 1

Đặt \(m=1-x=1-\frac{a+1}{a^2+a+1}=\frac{a^2+a+1-a-1}{a^2+a+1}=\frac{a^2}{a^2+a+1}\)

\(n=1-y=1-\frac{b+1}{b^2+b+1}=\frac{b^2+b+1-b-1}{b^2+b+1}=\frac{b^2}{b^2+b+1}\)

=>\(m:n=\frac{a^2}{a^2+a+1}:\frac{b^2}{b^2+b+1}\)

=>\(m:n=\frac{a^2}{a^2+a+1}.\frac{b^2+b+1}{b^2}\)

=>\(m:n=\frac{a^2.\left(b^2+b+1\right)}{\left(a^2+a+1\right).b^2}\)

=>\(m:n=\frac{a^2.b^2+a^2.b+a^2}{a^2.b^2+a.b^2+b^2}\)

=>\(m:n=\frac{a^2.b^2+ab.a+a^2}{a^2.b^2+ab.b+b^2}\)

Vì \(a>b=>ab.a>ab.b;a^2>b^2\)

=>\(a^2.b^2+ab.a+a^2>a^2.b^2+ab.b+b^2\)

=>\(\frac{a^2.b^2+ab.a+a^2}{a^2.b^2+ab.b+b^2}>1\)

=>m:n>1

=>m:n

=>1-x>y-y

=>x<y

Vậy x<y