BĐT $\Leftrightarrow \sqrt{\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-2}+\dfrac{8}{(\frac{a}{b}+1)(\frac{b}{c}+1)(\frac{c}{a}+1)}\geq 2$Lâu lắm rồi mới được vào đây
Bài toán 398.
Cho các số dương $a, b, c$. Chứng minh rằng :
$$\sqrt{\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-2}+\dfrac{8abc}{(a+b)(b+c)(c+a)}\ge 2$$
Đặt $\frac{a}{b}= x ,\frac{b}{c}= y,\frac{c}{a}= z\Rightarrow xyz=1$
BĐT $\Leftrightarrow \sqrt{x+y+z-2}+\dfrac{8}{(x+1)(y+1)(z+1)}\geq 2$
$\Leftrightarrow \sqrt{x+y+z-2}+\dfrac{8}{x+y+z+xy+yz+zx+2}\geq 2$
$\Leftrightarrow \sqrt{x+y+z-2}+\dfrac{8}{x+y+z+\frac{(x+y+z)^{2}}{3}+2}\geq 2$
Đặt tiếp $x+y+z=t\Rightarrow t\geq \sqrt[3]{xyz}= 3$
BĐT $\Leftrightarrow \sqrt{t-2}+\frac{8}{\frac{t^{2}}{3}+t+2}\geq 2\Leftrightarrow \sqrt{t-2}+\frac{24}{t^{2}+3t+6}\geq 2$
$\Leftrightarrow \Leftrightarrow \sqrt{t-2}-1\geq 1-\frac{24}{t^{2}+3t+6}\Leftrightarrow \frac{t-3}{\sqrt{t-2}+1}\geq \frac{(t-3)(t+6)}{t^{2}+3t+6}$
$\Leftrightarrow (t-3)(\frac{1}{\sqrt{t-2}+1}-\frac{t+6}{t^{2}+3t+6})\geq 0\Leftrightarrow \frac{1}{\sqrt{t-2}+1}-\frac{t+6}{t^{2}+3t+6}\geq 0$
$\Leftrightarrow t^{2}+3t+6\geq (\sqrt{t-2}+1)(t+6)\Leftrightarrow t^{2}+2t\geq (t+6)\sqrt{t-2}$
$\Leftrightarrow (t^{2}+2t)^{2}\geq [(t+6))\sqrt{t-2}]^{2}$
$\Leftrightarrow (t-3)(t^{3}+6t^{2}+12t+24)\geq 0$
Luôn đúng .
OK.