tui lm tắt thui nhak!
Nhân 2 xế vs $\sqrt{x+1}>0 $Đặt $\sqrt{x+1}=a$
BPT tg đg: $2x^2(x^2+1)+(x+2)a^2\geq a^5x+4ax$
$\Leftrightarrow 2x^2(x+1)^2-4x^3+(x+2)a^2\geq a^5x+4ax$
$\Leftrightarrow 2a^4x^2-4x^3+(x+2)a^2-a^5x-4ax\geq 0$
$\Leftrightarrow a^4x(2x-a)+x(a^2-4x^2)+2a(a-2x)\geq0 $
$\Leftrightarrow (a-2x)(-a^4x+a^3+a+2x^2)\geq 0$
$\Leftrightarrow (a-2x)(a^3+a-x^3-x)\geq 0$
$\Leftrightarrow (a-2x)(a-x)(a^2+ax+x^2+1)\geq 0$
~~ dễ rồi...