How do we know that $x^2 + \frac{1}{x^2}$ is greater or equal to $2$?

Since you tagged this with precalculus, we'll try the following. Start with the inequality $(x^2 - 1)^2 \geq 0$. Then, \begin{align*} (x^2 - 1)^2 \geq 0 && \implies && x^4 - 2x^2 + 1 &\geq 0 \\ && \implies && x^4 + 1 &\geq 2x^2 \\ && \implies && \frac{x^4 + 1}{x^2} & \geq 2 & \text{for } x \neq 0 \\ && \implies && x^2 + \frac{1}{x^2} & \geq 2. \end{align*} This was the inequality we wanted, except we have to be sure to exclude the case $x = 0$.

If $x\neq 0$ then $$\left(x-\frac{1}{x}\right)^2\ge0\iff x^2-2+\frac{1}{x^2}\ge 0 \iff x^2+\frac{1}{x^2}\ge 2$$

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