Cách khác:
Đặt \(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)
\(A=\left(1+\dfrac{a+b}{a}\right)\left(1+\dfrac{a+b}{b}\right)\)
\(A=\left(2+\dfrac{b}{a}\right)\left(2+\dfrac{a}{b}\right)\)
\(A=4+2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+1\)
\(A\ge4+2\cdot2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}+1=9\left(AM-GM\right)\left(đpcm\right)\)
( 1 + \(\dfrac{1}{a}\))\(\left(1+\dfrac{1}{b}\right)\) ≥ 9
Biến đổi VT Ta có : VT = \(\dfrac{a+1}{a}.\dfrac{b+1}{b}\)
= \(\dfrac{2a+b}{a}.\dfrac{2b+a}{b}\)
=\(\left(2+\dfrac{b}{a}\right)\left(2+\dfrac{a}{b}\right)\)
= 4 + \(\dfrac{2a}{b}+\dfrac{2b}{a}+\dfrac{b}{a}.\dfrac{a}{b}\)
= 5 + 2( \(\dfrac{a}{b}+\dfrac{b}{a}\) ) ( *)
Áp dụng BĐT : \(\dfrac{x}{y}+\dfrac{y}{x}\) ≥ 2( x > 0 ; y > 0) ( ** )
Từ ( * ; **) ⇒ 5 + 2( \(\dfrac{a}{b}+\dfrac{b}{a}\) ) ≥ 5 + 4 = 9 ( đpcm )
\(\text{Ta có : }\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\\ =1+\dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{ab}\\ =1+\dfrac{1}{ab}+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Với \(a;b>0\), áp dụng BDT Cô-si: \(\dfrac{\left(x+y\right)^2}{4}\ge xy\) và BDT: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}>0\\ \Rightarrow\dfrac{1}{ab}\ge\dfrac{1}{\dfrac{\left(a+b\right)^2}{4}}\\ \Rightarrow1+\dfrac{1}{ab}+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge1+\dfrac{1}{\dfrac{\left(a+b\right)^2}{4}}+\dfrac{4}{a+b}\\ \ge1+\dfrac{4}{1^2}+\dfrac{4}{1}\ge1+4+4\ge9\left(đpcm\right)\)
Vậy \(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\ge9\)
đẳng thức xảy ra khi \(a=b=\dfrac{1}{2}\)
