\(A=2x+\sqrt{4-2x^2}=\sqrt{2}.\sqrt{2x^2}+\sqrt{4-2x^2}\)
áp dụng BĐT bunhiacopxki,ta có:
\(A^2\le\left(2+1\right)\left(2x^2+4-2x^2\right)=3.4=12\)
\(\Leftrightarrow A\le\sqrt{12}\)
dấu = xảy ra khi \(\frac{\sqrt{2}}{\sqrt{2}x}=\frac{1}{\sqrt{4-2x^2}}\Leftrightarrow4-2x^2=x^2\Leftrightarrow x=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}\)
vậy Amax = \(\sqrt{12}\)khi x=\(\frac{2}{\sqrt{3}}\)