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?
 A: $x^2 + \frac{1}{x^2} = \frac{x^4 + 1}{x^2}$
$     = \frac{x^4 - 2x^2 + 1}{x^2} + 2$
$     = \frac{(x^2 - 1)^2}{x^2} + 2$
$     \geq 0 + 2 = 2$
A: Its not instantly but once you know it and have seen it a couple of times it becomes common sense. This is a consequence of Am-Gm inequality :a+b/2 ≥√(ab) substitute a=x² and b=(1/x)².
A: 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 $x = 0$.
A: You said imediately.  So this argument is worse than the others but it was the first thing I thought of.
this is equivalent to showing if $w \ge 1$ then $w + 1/w \ge 2$ which is equivalent to showing $w-1 \le 1 - 1/w $.
And $1 - 1/w = (w -1)/w \le w-1$
A: If $x\neq 0$ then $$\left(x-\frac{1}{x}\right)^2\ge0\iff x^2-2+\frac{1}{x^2}\ge 0 \iff x^2+\frac{1}{x^2}\ge 2$$
A: 
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A: One way is using the arithmetic-geometric mean inequality: $(A+B)/2\ge \sqrt{AB}$
for $A\ge 0, B\ge 0$. Take $A=x^2$ and $B=1/x^2$.
