ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\\frac{-1-\sqrt{5}}{4}\le x\le-\frac{1}{8}\end{matrix}\right.\)(Có thể chưa chính xác)
\(12x^2+16x+1=2\sqrt{24x^3+12x^2-6x}+4\sqrt{x^2-x}+4\sqrt{8x^3+9x^2+x}\)
Áp dụng AM-GM:
\(2\sqrt{24x^3+12x^2-6x}=2\sqrt{6x\left(4x^2+2x-1\right)}\le6x+\left(4x^2+2x-1\right)=4x^2+8x-1\left(1\right)\)
\(4\sqrt{x^2-x}=2\sqrt{1.\left(4x^2-4x\right)}\le4x^2-4x+1\left(2\right)\)
\(4\sqrt{8x^3+9x^2+x}=2\sqrt{\left(4x^2+4x\right)\left(8x+1\right)}\le\left(4x^2+4x\right)+\left(8x+1\right)=4x^2+12x+1\left(3\right)\)
Cộng \(\left(1\right),\left(2\right),\left(3\right)\), ta có: \(VP\le VT\)
Dấu ''='' xảy ra khi :
\(\left\{{}\begin{matrix}4x^2+2x-1=6x\\4x^2-4x=1\\4x^2+4x=8x+1\end{matrix}\right.\)\(\Rightarrow4x^2-4x-1=0\)
\(\Rightarrow x=\frac{1\pm\sqrt{2}}{2}\) (t/m ĐKXĐ)
