Đáp án+Giải thích các bước giải:
$\begin{array}{l}\text{áp dụng BĐT cosi với 2 số dương ta có} \dfrac{1}{x^2}+\dfrac{1}{y^2} \geq 2\dfrac{1}{xy}\\CMTT:\dfrac{1}{y^2}+\dfrac{1}{z^2} \geq 2\dfrac{1}{yz}\\\dfrac{1}{x^2}+\dfrac{1}{z^2} \geq 2\dfrac{1}{xz}\\↔2(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}) \geq 2(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz})\\↔\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2} \geq \dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\\↔\dfrac{1}{x^2}+\dfrac{1}{y^2}+(\dfrac{1}{z^2}+2(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}) \geq 3(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz})\\↔(\dfrac{1}{x}+\dfrac{1}{y}+(\dfrac{1}{z})^2 \geq 3(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz})\\↔(\dfrac{xyz}{x}+\dfrac{xyz}{y}+(\dfrac{xyz}{z})^2 \geq 3(\dfrac{xyz}{xy}+\dfrac{xyz}{yz}+\dfrac{xyz}{xz})\\↔(xy+yz+zx)^2 \geq 3(x+y+z)\\↔xy+yz+zx \geq \sqrt{3(x+y+z)}\\↔\dfrac{1}{xy+yz+zx} \leq \dfrac{1}{\sqrt{3(x+y+z)}}\\↔2\dfrac{1}{xy+yz+zx} \leq \dfrac{2}{\sqrt{3(x+y+z)}}\\↔-2\dfrac{1}{xy+yz+zx} \geq -\dfrac{2}{\sqrt{3(x+y+z)}}\\↔A \geq \dfrac{1}{x+y+z}-\dfrac{2}{\sqrt{3(x+y+z)}}\\↔A \geq \dfrac{1}{x+y+z}-2.\dfrac{1}{\sqrt{x+y+z}}.\dfrac{1}{\sqrt{3}}+\dfrac{1}{3}-\dfrac{1}{3}\\↔A \geq (\dfrac{1}{\sqrt{x+y+z}}-\dfrac{1}{\sqrt{3}})^2-\dfrac{1}{3} \geq -\dfrac{1}{3}\\\text{dấu = xảy ra khi x=y=z=1}\\\text{vậy GTNN_A=-1/3↔x=y=z=1} \\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
