Nhận xét : A > 0
Áp dụng bđt Bunhiacopxki , ta có :
\(A^2=\left(1.\sqrt{x-1}+1.\sqrt{9-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+9-x\right)\)
\(\Rightarrow A^2\le16\Rightarrow A\le4\)
Suy ra Max A = 4 <=> \(\begin{cases}1\le x\le9\\\sqrt{x-1}=\sqrt{9-x}\end{cases}\) \(\Leftrightarrow x=5\)
\(A^2=\left(x-1\right)+\left(9-x\right)+2\sqrt{\left(x-1\right)\left(9-x\right)}\)
\(=8+2\sqrt{\left(x-1\right)\left(9-x\right)}\).Dùng BĐT cô-si
\(=8+2\sqrt{\left(x-1\right)\left(9-x\right)}\le8+\left(x-1\right)+\left(9-x\right)=16\)
\(\Rightarrow A^2\le16\Leftrightarrow A\le4\)
Dấu = khi \(\begin{cases}1\le x\le9\\\sqrt{x-1}=\sqrt{9-x}\end{cases}\)\(\Leftrightarrow x=5\)
Vậy MaxA=4 khi x=5
