$\begin{array}{l}\rm Câu\,\,1:\\ \quad A = \dfrac{x^2 + 2x + 3}{(x+2)^2}\qquad (x\ne -2)\\ \to A = \dfrac{3x^2 + 6x + 9}{3(x+2)^2}\\ \to A = \dfrac{2x^2 + 8x + 8 + x^2 - 2x + 1}{3(x+2)^2}\\ \to A = \dfrac{2(x+2)^2 + (x-1)^2}{3(x+2)^2}\\ \to A = \dfrac23 + \dfrac{(x-1)^2}{3(x+2)^2} \\ \text{Ta có:}\\ \quad (x-1)^2 \geq 0\quad \forall x\\ \to \dfrac{(x-1)^2}{3(x+2)^2} \geq 0\quad \forall x \ne -2\\ \to \dfrac23 + \dfrac{(x-1)^2}{3(x+2)^2} \geq \dfrac23\\ \to A \geq \dfrac23\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1\\ Vậy\,\,\min A = \dfrac23 \Leftrightarrow x =1\\ \rm Câu\,\,2:\\ Với\,\,a;b;c>0\\ \text{Áp dụng bất đẳng thức $AM-GM$ ta được:}\\ \quad a + b+ c \geq 3\sqrt[3]{abc}\\ \quad \dfrac1a + \dfrac1b + \dfrac1c \geq 3\sqrt[3]{\dfrac{1}{abc}}\\ \text{Nhân vế theo vế ta được:}\\ \quad (a + b+ c)\left(\dfrac1a + \dfrac1b + \dfrac1c\right) \geq 3\sqrt[3]{abc}\cdot 3\sqrt[3]{\dfrac{1}{abc}}\\ \to (a + b+ c)\left(\dfrac1a + \dfrac1b + \dfrac1c\right) 9\sqrt[3]{abc\cdot\dfrac{1}{abc}}\\ \to (a + b+ c)\left(\dfrac1a + \dfrac1b + \dfrac1c\right) \geq 9\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow a =b = c\end{array}$