\(D=\sqrt{9-x}+\sqrt{x}\)
Áp dụng BĐT Bunhiacopxki, ta có:
\(\left(\sqrt{9-x}+\sqrt{x}\right)^2\le\left(1^2+1^2\right)\left[\left(\sqrt{9-x}\right)^2+\left(\sqrt{x}\right)^2\right]\)
\(\Leftrightarrow D^2\le2.\left(9-x+x\right)\)
\(\Leftrightarrow D^2\le18\)
\(\Leftrightarrow D\le3\sqrt{2}\)
Dấu "=" xảy ra khi \(\dfrac{1}{\sqrt{9-x}}=\dfrac{1}{\sqrt{x}}\)
\(\Leftrightarrow\sqrt{9-x}=\sqrt{x}\)
\(\Leftrightarrow x=\dfrac{9}{2}\)
Vậy \(Max_D=3\sqrt{2}\) khi \(x=\dfrac{9}{2}\)
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