Prove that $-\sqrt{c}<ab<0$ if $a^4-2019a=b^4-2019b=c$.

Notice that, since $a\ne b$ $$a^4-b^4= 2019(a-b)\implies (a^2+b^2)(a+b)=\color{red}{2019}$$

If we add both equations we get:

\begin{align}2c &= a^4+b^4-\color{red}{2019}(a+b) \\ &= a^4+b^4-(a+b)^2(a^2+b^2)\\ &= -2ab(a^2+ab+b^2) \end{align} Since $a^2+ab+b^2>0$ we have $ab<0$. Further

\begin{align}c &= -ab(a^2+ab+b^2)\\ &> -ab(2|ab|+ab)\\ &= -ab(-2ab+ab)\\ & = a^2b^2 \end{align}

So $c>a^2b^2$ which means $-\sqrt{c}<ab$