mình bổ sung thêm đề: a,b dương
BÀI LÀM
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\)
\(=\left(1+\frac{a+b}{a}\right)\left(1+\frac{a+b}{b}\right)\) (thay a+b = 1)
\(=\left(1+\frac{a}{a}+\frac{b}{a}\right)\left(1+\frac{a}{b}+\frac{b}{b}\right)\)
\(=\left(2+\frac{b}{a}\right)\left(2+\frac{a}{b}\right)\)
\(=4+2\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{b}{a}.\frac{a}{b}\)
\(=5+2\left(\frac{a}{b}+\frac{b}{a}\right)\) \(\ge5+2.2=9\) (1)
c/m: \(\frac{a}{b}+\frac{b}{a}\ge2\) với a,b dương
\(\Leftrightarrow\) \(\frac{a^2}{ab}+\frac{b^2}{ab}\ge\frac{2ab}{ab}\)
\(\Leftrightarrow\)\(\frac{a^2}{ab}+\frac{b^2}{ab}-\frac{2ab}{ab}\ge0\)
\(\Leftrightarrow\)\(\frac{\left(a-b\right)^2}{ab}\ge0\) luôn đúng
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b\)
Vậy BĐT (1) đã được chứng minh
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=\frac{1}{2}\)
Theo Cauchy , ta có \(a+b\ge2\sqrt{ab}\Rightarrow\sqrt{ab}\le\frac{a+b}{2}\)
Áp dụng bất đẳng thức Bunyakovsky , ta có :
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge\left(1+\frac{1}{\sqrt{a}.\sqrt{b}}\right)^2\ge\left(1+\frac{1}{\frac{\left(a+b\right)}{2}}\right)^2=\left(1+2\right)^2=9\)
Đẳng thức xảy ra <=> a = b = 1/2
