Proof of $x+\sqrt{x^2+e} >0~~~~\text{where}~~x\in[-1,1]$

Question

$$e:=\exp(1)\tag{1}$$

I want to prove the following positivity.

$$x+\sqrt{x^2+e}　>0~~~~\text{where}~~x\in[-1,1]\tag{2}$$

I've tried to prove it using proof of contradiction.

$$\underbrace{x+\sqrt{x^2+e}<0}_{{\text{Assumption}}}\tag{3}$$

$$\underbrace{\sqrt{x^2+e}}_{\text{positive}}<-x~~\implies~~-1\leq x<0\tag{4}$$

Above RHS inequalities mean that all section$~[-1,1]~$is not covered hence eqn3 is wrong$~~\Leftrightarrow~~$$~x+\sqrt{x^2+e}>0~$is true.

But my this statement is too rough since I didn't specify that$~\sqrt{x^2+e}<-x~$can be satisfied for all$~-1\leq x<0~$

Any more wise way to prove line2?

Answer 1

$\sqrt {x^{2}+e} \geq \sqrt e >1\geq -x$.

Answer 2

Alternative approach:

If $r$ is any positive value, and $x$ is any real number, then $\sqrt{x^2 + r} > \sqrt{x^2} = |x| \geq \pm x.$