\(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)
\(A=1+\dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{ab}\)
\(A=1+\dfrac{a+b}{ab}+\dfrac{a+b}{ab}\)
\(A=1+\dfrac{2}{ab}\)
Ta có:\(\left(a-b\right)^2\ge0\)
\(\Rightarrow a^2+b^2\ge2ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow ab\le\dfrac{1}{4}\)
\(\Rightarrow A\ge1+\dfrac{2}{\dfrac{1}{4}}=9\)
"="<=>a=b=0,5
Áp dụng liên tiếp bất đẳng thức CauChy-Schwarz và AM-GM\(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)
\(A=1+\dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{ab}\)
\(A\ge1+\dfrac{\left(1+1\right)^2}{a+b}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{4}}\)
\(A\ge1+\dfrac{4}{a+b}+\dfrac{4}{\left(a+b\right)^2}=1+4+4=9\)
Dấu "=" xảy ra khi: \(a=b=\dfrac{1}{2}\)
