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?

What you can show is already stronger than what you want to show: $|x| \le 2|x|$. — Erick Wong, Apr 16 '17 at 07:12
This can be also considered as a special case of $xy\le x^2+y^2$, see How can I prove that $xy\leq x^2+y^2$? and other questions linked there. — Martin Sleziak, Aug 29 '19 at 07:07
Here is a link to the discussion of this question in the reopen request thread. — Martin Sleziak, Aug 29 '19 at 07:15
For people who are wondering how to get the inequality $2|x|\le 1+x^2$ mentioned in the post, see here (for example): How to prove that $x^2 +1 \geq 2x$? — Martin Sleziak, Aug 29 '19 at 13:50

score 6 · Accepted Answer · answered Apr 16 '17 at 07:08

6

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

answered Apr 16 '17 at 07:08

Olivier Oloa

120,989

score 2 · Answer 2 · answered Apr 16 '17 at 07:09

2

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

answered Apr 16 '17 at 07:09

lab bhattacharjee

274,582

score 2 · Answer 3 · answered Apr 16 '17 at 07:09

2

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 :

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

answered Apr 16 '17 at 07:09

Jaideep Khare

19,293

Though in the section right after "You get" both implications are true, technically the one you are interested in is "$\Longleftarrow$", not "$\implies$". – Apr 16 '17 at 07:23
@G.Sassatelli I think there should be a bi-implies arrow, instead of a single sided arrow.Right? – Jaideep Khare Apr 16 '17 at 07:25
Yes, it is true that the two inequations are equivalent. My point is that you've stated only the implication which does not serve your purpose, since you want to use the fact that the second is always true to prove that the first is always true. – Apr 16 '17 at 07:27
@G.Sassatelli Yes, of couse they are. – Jaideep Khare Apr 16 '17 at 07:28

score 2 · Answer 4 · answered Apr 16 '17 at 07:13

2

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

answered Apr 16 '17 at 07:13

The way of life

1,192

What do you mean by 'inserting the inequality'? – chris Apr 16 '17 at 08:21
The transitive rule. – The way of life Apr 16 '17 at 08:39

score 2 · Answer 5 · answered Apr 16 '17 at 07:19

2

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

answered Apr 16 '17 at 07:19

Dr. Sonnhard Graubner

95,283

score 0 · Answer 6 · answered Apr 16 '17 at 08:10

0

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.

answered Apr 16 '17 at 08:10

fleablood

124,253

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

6 Answers6