Nhầm min not max:v
\(A=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=1+1+\dfrac{a}{b}+\dfrac{b}{a}\ge1+1+2=4\)
"=" Khi a=b
Áp dụng BĐT Cô - Si dạng Engel , có :
\(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\)
⇒ \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) ≥ 4
⇒ AMIN = 4 ⇔ a = b
P/S : Tìm GTNN nhé ( sai đề rùi )
\(A=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(A=1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
\(A=\dfrac{a}{b}+\dfrac{b}{a}+2\)
Áp dụng BĐT Cauchy ( ko có engle) cho 2 số không âm
Ta có: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
=> \(A=2+\dfrac{a}{b}+\dfrac{b}{a}\ge2+2=4\)
MinA = 4 khi a = b
