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

Question

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

I can show $2|x|\leq 1+x^2$, but how to show the above?

Answer 1

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

Answer 2

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

Answer 3

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 :

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

Answer 4

Since $|x|\geq 0$, you have $|x|\leq 2|x|$, and so by inserting the inequality you know $2|x|\leq 1+x^2$ you'd get the desires $|x|\leq 2|x|\leq 1+x^2$.

Answer 5

by $AM-GM$ we have $$1+x^2\geq 2|x|\geq |x|$$

Answer 6

Well you've actually proven something stronger.

Since $|x|\ge 0$ and $2>1$, then $|x|\le 2|x| \le 1 + x^2$.

But, if you want to be thorough.

$(|x|-1)^2\ge 0$

$x^2-2|x|+1\ge 0$

$x^2 +1\ge 2|x|$

$x^2+1\ge 2|x| \ge |x|$.

That's all.