Prove that $a^4+b^4+1\ge a+b$.

Note that, by the AM-GM Inequality, $$x^4+\frac{1}{2}=x^4+3\left(\frac{1}{6}\right)\geq 4\sqrt[4]{x^4\left(\frac{1}{6}\right)^3}=\sqrt[4]{\frac{256}{216}}\,|x|\geq |x|\geq x$$ for all $x\in\mathbb{R}$. The equality does not hold, though.

---

A sharper inequality is $a^4+b^4+\frac{3}{\sqrt[3]{2}^5}\geq a+b$. This is because $$x^4+\frac{3}{\sqrt[3]{2}^8} =x^4+3\left(\frac{1}{\sqrt[3]{2}^8}\right)\geq 4\sqrt[4]{x^4\left(\frac{1}{\sqrt[3]{2}^8}\right)^3}=|x|\geq x\,.$$ The inequality above is an equality if and only if $x=\frac{1}{\sqrt[3]{2}^2}$.

Proof

Since $$\left(a^4-a^2+\frac{1}{4}\right)+\left(b^4-b^2+\frac{1}{4}\right)=\left(a^2-\frac{1}{2}\right)^2+\left(b^2-\frac{1}{2}\right)^2 \geq0,$$ hence $$a^4+b^4+1 \geq a^2+b^2+\frac{1}{2}.\tag1$$

Since $$\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)=\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2 \geq 0,$$ hence, $$a^2+b^2+\frac{1}{2} \geq a+b.\tag2$$

Combining $(1)$ and $(2)$, $$a^4+b^4+1 \geq a+b.$$

We have

$$a^4+b^4+1\ge a+b\iff a^4-a+b^4-b+1\ge 0$$

and it is easy to show that

$$f(x)=x^4-x\implies f'(x)=4x^3-1\implies x_{min}=\frac1{4^\frac13}\quad f(x)\ge f(x_{min})\approx-0.4725$$