Question

cmr:\(x^2+3+\dfrac{1}{x^2+3}\ge\dfrac{10}{3}\)

Answer 1

Lời giải:

Ta có:

$A=x^2+3+\frac{1}{x^2+3}=\frac{x^2+3}{9}+\frac{1}{x^2+3}+\frac{8(x^2+3)}{9}$

Thấy rằng:

$\frac{8(x^2+3)}{9}\geq \frac{8(0+3)}{9}=\frac{8}{3}(1)$ (do $x^2\geq 0$ với mọi $x$)

$\frac{x^2+3}{9}+\frac{1}{x^2+3}\geq 2\sqrt{\frac{x^2+3}{9}.\frac{1}{x^2+3}}=\frac{2}{3}(2)$

Từ $(1); (2)\Rightarrow A\geq \frac{8}{3}+\frac{2}{3}=\frac{10}{3}$ (đpcm)

Dấu '=' xảy ra khi $x=0$

Answer 2

Lời giải:

Ta có:

$A=x^2+3+\frac{1}{x^2+3}=\frac{x^2+3}{9}+\frac{1}{x^2+3}+\frac{8(x^2+3)}{9}$

Thấy rằng:

$\frac{8(x^2+3)}{9}\geq \frac{8(0+3)}{9}=\frac{8}{3}(1)$ (do $x^2\geq 0$ với mọi $x$)

$\frac{x^2+3}{9}+\frac{1}{x^2+3}\geq 2\sqrt{\frac{x^2+3}{9}.\frac{1}{x^2+3}}=\frac{2}{3}(2)$

Từ $(1); (2)\Rightarrow A\geq \frac{8}{3}+\frac{2}{3}=\frac{10}{3}$ (đpcm)

Dấu '=' xảy ra khi $x=0$