Prove that $a^4+b^4+1\ge a+b$. 
Prove that
$$a^4+b^4+1\ge a+b$$
for all real numbers $a,b$.

What I've tried:
1.I checked how AM-GM may help but doesn't look like it's useful here.


*I've tried:

$$(a^2+b^2)^2 -2(ab)^2+1 \ge a+b$$
But unfortunately, I can't find any way to continue this..
I'm sure that this isn't too hard, it's just that "I'm not seeing it", I would appreciate if clues are given first so I can answer this myself.
*This exercise is from the TAU entry exams.
 A: 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
$$f(x)=x^4-x\implies f'(x)=4x^3-1\implies x_{min}=\frac1{4^\frac13}\quad f(x)\ge f(x_{min})\approx-0.4725$$
A: Clues only in this answer, since you said you want to figure out the answer yourself.
Have you tried proving it for positive numbers greater than $2$? The square of such a number is obviously greater than $2$, and much more so the biquadrate. Prove it for those numbers and the answer is obvious for negative numbers less than $-2$.
Where it gets really tricky is in the unit interval. Let's say $$a = b = \frac{1}{2}.$$ Then $$\left(\frac{1}{2}\right)^4 = \frac{1}{16},$$ so $$a^4 + b^4 = \frac{1}{8},$$ which is actually less than either $a$ or $b$ alone. But then you get to add $1$ to that...
A: Let's have $4c^3=1$ and $f(a)=a^4-a+\frac 12$
$\begin{align}\require{cancel}f(c+u)-f(c) &=\cancel{c^4}+\cancel{4c^3u}+6c^2u^2+4cu^3+u^4-\cancel{c}-\cancel{u}+\cancel{\frac 12}-\cancel{c^4}+\cancel{c}-\cancel{\frac 12}\\
&=u^2\underbrace{(u^2+4cu+6c^2)}_{\Delta=-8c^2<0}\ge 0\end{align}$
So $f(c)$ is a minimum and $f(a)\ge f(c)\approx0.0275>0$
The conclusion arises from $f(a)+f(b)>0$.
A: 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 $x=\frac{1}{\sqrt[3]{2}^2}$.
A: 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 $(2)$, $$a^4+b^4+1 \geq a+b.$$
