Có: \(\left(x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}\right)^2\ge\left(x^2+12-x^2\right)\left(12-y+y\right)=12^2\)(Bunhiacopxki)
\(\Rightarrow x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}\ge12\)
Dấu "=" xảy ra <=> \(\frac{x}{\sqrt{12-y}}=\frac{\sqrt{12-x^2}}{\sqrt{y}}\)\(\Leftrightarrow\frac{x^2}{12-y}=\frac{12-x^2}{y}=\frac{x^2+12-x^2}{12-y+y}=1\)
\(\Rightarrow x^2=12-y\Rightarrow y=12-x^2\)
Có :\(x^3-8x-1=2\sqrt{12-x^2-2}=2\sqrt{10-x^2}\)