$\sqrt{2x^3+4x^2+4x}-\sqrt[3]{16x^3+12x^2+6x-3}\geq 4x^4+2x^3-2x-1$

Bài toán: 
Giải bất phương trình:
$$\sqrt{2x^3+4x^2+4x}-\sqrt[3]{16x^3+12x^2+6x-3}\geq 4x^4+2x^3-2x-1$$




Lời giải:ĐK: $2x(x^2+2x+2)\geq 0\Leftrightarrow x\geq 0$
$$BPT\Leftrightarrow \sqrt{2x^3+4x^2+4x}-(2x+1)+(2x+1)-\sqrt[3]{16x^3+12x^2+6x-3}$$
$$\geq (2x^3-1)(2x+1)$$
Đặt $\left\{\begin{matrix}a=\sqrt{2x^3+4x^2+4x}-(2x+1)  &  & \\ A=\sqrt{2x^3+4x^2+4x}+(2x+1)  &  & \\ b=(2x+1)-\sqrt[3]{16x^3+12x^2+6x-3} \\ B=(2x+1)^2+(2x+1)\sqrt[3]{16x^3+12x^2+6x-3}+\sqrt[3]{(16x^3+12x^2+6x-3)^2} \end{matrix}\right.$

$\Leftrightarrow \left\{\begin{matrix}a.A=2x^3-1  &  & \\ b.B=-8x^3+4=-4(2x^3-1)  &  &  \end{matrix}\right.$
$BPT$ trở thành:
$$a+b\geq (2x^3-1)(2x+1)$$
$$\Leftrightarrow \dfrac{a.A}{A}+\dfrac{b.B}{B}-(2x^3-1)(2x+1)\geq 0$$
$$\Leftrightarrow \dfrac{2x^3-1}{A}-\dfrac{4(2x^3-1)}{B}-(2x^3-1)(2x+1)\geq 0$$
$$\Leftrightarrow (2x^3-1)\left (\dfrac{1}{A}-\dfrac{4}{B}-2x-1  \right )\geq 0$$
$$\Leftrightarrow (2x^3-1)\left (\dfrac{1-A}{A}-\dfrac{4}{B}-2x  \right )\geq 0$$
Do $x\geq 0\Rightarrow A\geq 1$ và $B>0$ (bình phương thiếu một tổng)
$$\Rightarrow \dfrac{1-A}{A}<0;\dfrac{-4}{B}<0;-2x\leq 0$$
$$\Rightarrow \left (\dfrac{1-A}{A}-\dfrac{4}{B}-2x  \right )\leq 0$$
$$\Rightarrow 2x^3-1\leq 0$$
$$\Leftrightarrow 0\leq x\leq \dfrac{1}{\sqrt[3]{2}}=\dfrac{\sqrt[3]{4}}{2}$$