How to show for $x\in \mathbb{R}$, $|x|\leqslant 1+x^2$

Observe that $$ 2|x|\leq 1+x^2 \quad \implies \quad |x|\leq \frac{1+x^2}2 \le 1+x^2. $$

For real $x,|x|^2=x^2$ and $$4|x|^2-4|x|+4=(2|x|-1)^2+3\ge3$$

Let $|x|=t$ ;

You get : $$|x|\leqslant 1+x^2 \implies t^2-t+1 \ge 0$$

Now, the discriminant of $t^2-t+1$ is : $$b^2-4ac=1-4=-3 <0$$ Therefore;

$$ t^2-t+1 \ge 0 ~\forall ~t \in \mathbb R^+$$ Hence :

$$|x|\leqslant 1+x^2 ~~\forall ~x \in \mathbb R$$