How do we know that $x^2 + \frac{1}{x^2}$ is greater or equal to $2$? For one problem, we were supposed to know that: 
$$x^2 + \frac{1}{x^2}\geq 2.$$
How do you deduce this instantly when looking at the expression above?

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 $x = 0$.
