\(\frac{\sqrt{ax+1}\left(\sqrt[3]{bx+1}-1\right)+\sqrt{ax+1}-1}{x}=\frac{\frac{bx\sqrt{ax+1}}{\sqrt[3]{\left(bx+1\right)^2}+\sqrt[3]{bx+1}+1}+\frac{ax}{\sqrt{ax+1}+1}}{x}=\frac{b\sqrt{ax+1}}{\sqrt[3]{\left(bx+1\right)^2}+\sqrt[3]{bx+1}+1}+\frac{a}{\sqrt{ax+1}+1}\)
\(\Rightarrow\lim\limits_{x\rightarrow0}f\left(x\right)=a+b\Rightarrow a+b=1\)
\(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2=\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)