Proving the integral inequality $2≤\int_{-1}^1 \sqrt{1+x^6} \,dx ≤ 2\sqrt{2} $

So there are two inequalities to be proved. You can use that $\sqrt{1+x^6} \leq \sqrt{2}$ for all $x \in [-1,1]$ for the upper bound, as it follows $\int_{[-1,1]} \sqrt{1+x^6} dx\leq \int_{[-1,1]} \sqrt{2} dx\leq 2 \sqrt{2}$. The lower bound follows very similarly.

$ \int_{-1}^1 \sqrt{1+x^6}dx=2\int_{0}^1 \sqrt{1+x^6} dx$ because $\sqrt{1+x^6}$ is an even function. so we must show: $$2≤2\int_{0}^1 \sqrt{1+x^6} dx ≤ 2\sqrt{2}$$ or we must show: $$1≤\int_{0}^1 \sqrt{1+x^6} dx ≤ \sqrt{2}$$

$1≤\sqrt{1+x^6}$ then $\int_{0}^11dx\leq\int_{0}^1 \sqrt{1+x^6} dx$ we have $1≤\int_{0}^1 \sqrt{1+x^6} dx$ $$$$

we have $1+x^6\leq2$ if $0\leq x \leq1$ and then we have $\sqrt{1+x^6}\leq \sqrt2$ if $0\leq x \leq1$ therefore : $$\int_{0}^1\sqrt{1+x^6}dx\leq \int_{0}^1\sqrt2dx=\sqrt2$$

[Edited after the meta question and this conversation.]

Let $f(x)=\sqrt{1+x^6}$. It is evident that the argument of the square root $1+x^6\ge 1\ \forall\ x\in\mathbb R$ and there is only one point of global minima $(0, 1) $. Hence, $f(x)$ is monotically decreasing and increasing for $x\le 0$ and $x\ge 0$ respectively.

$$\text{Area}(\square CEFD)\le \text{Area under curve} \le \text{Area}(\square ABDC) \\ \implies 2\le \int_{-1}^{1} \sqrt{1+x^6}\ dx\le 2\sqrt 2$$