Giải thích các bước giải:
Ta có:
$A=x^2+\dfrac{1}{(x^2+1)^2}$
$\to A=\dfrac12(x^2+1)+\dfrac12(x^2+1)+\dfrac{1}{(x^2+1)^2}-1$
$\to A\ge 3\sqrt[3]{\dfrac12(x^2+1)\cdot\dfrac12(x^2+1) \cdot\dfrac{1}{(x^2+1)^2}}-1$
$\to A\ge 3\sqrt[3]{\dfrac14}-1$
$\to GTNN_A=3\sqrt[3]{\dfrac14}-1$
Khi đó $\dfrac12(x^2+1)=\dfrac{1}{(x^2+1)^2}$
$\to x=\pm\sqrt{\sqrt[3]{2}-1}$