Ta có :
$a+b\ge1$
$⇔b\ge1-a$
$⇔ A\ge\dfrac{8a^2+1-a}{4a}+\left(1-a\right)^2=a^2+\dfrac{1}{4a}+\dfrac{3}{4}=a^2+\dfrac{1}{8a}+\dfrac{1}{8a}+\dfrac{3}{4}$Ta áp dụng BĐT Cauchy :
$⇒a^2+\dfrac{1}{8a}+\dfrac{1}{8a}\ge3\sqrt[3]{a^2.\dfrac{1}{8a}.\dfrac{1}{8a}}=\dfrac{3}{4}$
$⇒ A\ge\dfrac{3}{2}$
`⇒ A_min = 3/2`
Dấu "=" xảy ra khi $a^2=\dfrac{1}{8a}$
$⇔ a=\dfrac{1}{2}$
$⇔ b=\dfrac{1}{2}$
Vậy `A_min=3/2` khi `a=b=1/2`
Học tốt !
