Chủ đề này có 1 trả lời

[Tahsi]

(x^3+2x^2+x)/(x^4+2x^2+1)

[DOTOANNANG]

Đặt $\mathrm{A}\left ( x \right )= \frac{x^{\,3}+ 2\,x^{\,2}+ x}{x^{\,4}+ 2\,x^{\,2}+ 1}$

$${\mathrm{A}}'\left ( x \right )= \frac{\mathrm{d} }{\mathrm{d} \mathrm{A}}\,\left ( \frac{x^{\,3}+ 2\,x^{\,2}+ x}{x^{\,4}+ 2\,x^{\,2}+ 1} \right )= \frac{\left ( 1- x \right )\left ( 1+ x \right )\left ( x^{\,2}+ 4\,x+ 1 \right )}{\left ( x^{\,2}+ 1 \right )^{\,3}}= 0\,\,\Leftrightarrow \,\,x= 1,\,x= -\,1,\,x= -\,\frac{x^{\,2}+ 1}{4}$$

Vậy, ta có bất đẳng thức chặt hơn: $-\,\frac{1}{8}= \mathrm{A}\left ( -\,2- \sqrt{3} \right )= \mathrm{A}\left ( \sqrt{3}- 2 \right )\leqq \mathrm{A}\left ( -\frac{x^{\,2}+ 1}{4} \right )\leqq \mathrm{A}\left ( x \right )\leqq \mathrm{A}\left ( 1 \right )= 1$

Spoiler

$\mathrm{A}\left ( x \right )+ \frac{1}{8}= \frac{\left ( x^{\,2}+ 4\,x+ 1 \right )^{2}}{8\left ( x^{\,2}+ 1 \right )^{2}}\geqq 0,\,\,1- \mathrm{A}\left ( x \right )= \frac{\left ( x- 1 \right )^{\,2}\left ( x^{\,2}+ x+ 1 \right )}{\left ( x^{\,2}+ 1 \right )^{\,2}}\geqq 0$