Áp dụng BĐT Cauchy ta có:
$\sqrt[3]{4x(8x+1)}=\dfrac{1}{6}.3\sqrt[3]{2.16x.(8x+1)}\le\dfrac{2+16x+8x+1}{6}=4x+\dfrac{1}{2}$
$\Rightarrow 96x^2-20x+2\le4x+\dfrac{1}{2}$
$\Leftrightarrow 96x^2-24x+\dfrac{3}{2}\le0$
$\Leftrightarrow \dfrac{3}{2}(8x-1)^2\le0$
$\Leftrightarrow x=\dfrac{1}{8}$, thỏa mãn.