<s>[INST] 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.
 [/INST] 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 $(2)$, $$a^4+b^4+1 \geq a+b.$$
</s>