TITLE: Prove that $a^4+b^4+1\ge a+b$.
QUESTION [11 upvotes]: 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.

REPLY [2 votes]: 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 $f(a)+f(b)>0$.