Junior Balkan Mathematical Olympiad

Let $a^4 - 2019a = b^4 - 2019b = c$.

Since $a \neq b$, subtract the two equations:
$$a^4 - b^4 - 2019(a - b) = 0$$
$$(a - b)(a^3 + a^2b + ab^2 + b^3) - 2019(a - b) = 0$$
Since $a \neq b$, divide both sides by $(a - b)$:
$$a^3 + a^2b + ab^2 + b^3 = 2019$$
$$(a + b)^3 - 3ab(a + b) = 2019$$
Let $s = a + b$, $p = ab$.
Then:
$$s^3 - 3p s = 2019$$
So:
$$s^3 - 3p s - 2019 = 0$$
Also, from $a^4 - 2019a = c$:
$$a^4 - 2019a - c = 0$$
Similarly for $b$:
$$b^4 - 2019b - c = 0$$
So $a$ and $b$ are roots of the quartic $x^4 - 2019x - c = 0$.
Let $x^4 - 2019x - c = 0$ have roots $a$, $b$, $r$, $s$.
By Vieta's formulas:
- $a + b + r + s = 0$
- $ab + ar + as + br + bs + rs = 0$
- $abc + abr + abs + ars + brs + bsr + rsr = 2019$
- $abrs = -c$

But we only need $a$ and $b$.

Let us consider the function $f(x) = x^4 - 2019x - c$.
Since $c > 0$, for large $|x|$, $f(x) > 0$.
Let us check the sign of $f(x)$ at $x = 0$:
$$f(0) = -c < 0$$
So $f(x)$ has at least two real roots, one positive and one negative.
Let $a > 0$, $b < 0$.

Now, $ab < 0$.

Let us show that $ab > -\sqrt{c}$.

Suppose $a > 0$, $b < 0$.
Then $a^4 - 2019a = b^4 - 2019b = c$.

Let $a > 0$, $b < 0$.
Then $a^4 > 2019a + c > c$ (since $a > 0$).
Similarly, $b^4 < 2019b + c < c$ (since $b < 0$).
But $b^4$ is positive, $2019b$ is negative, so $b^4 - 2019b$ increases as $b$ becomes more negative.

Let us try to bound $ab$.

Let $a$ and $b$ be roots of $x^4 - 2019x - c = 0$.
Then $x^4 - 2019x = c$.
Let $x = t$, $t^4 - 2019t = c$.

Let $a > 0$, $b < 0$.
Then $ab < 0$.

Now, consider $a^4 - 2019a = b^4 - 2019b$.
Let $a = -b$.
Then $a^4 - 2019a = b^4 - 2019b$ becomes $a^4 - 2019a = a^4 + 2019a$, so $-2019a = 2019a$, so $a = 0$, but $a \neq b$.
So $a \neq -b$ unless $a = 0$.

Now, since $a$ and $b$ are distinct, and $a > 0$, $b < 0$, $ab < 0$.

Let us try to bound $ab$ from below.

Let $a^4 - 2019a = c$.
Then $a^4 = 2019a + c$.
So $a^4 \geq c$ for $a > 0$.
So $a \geq c^{1/4}$.

Similarly, for $b < 0$, $b^4 = 2019b + c$.
But $b^4$ is positive, $2019b$ is negative, so $b^4 < c$ for $b$ negative and large in magnitude.

Now, $ab$ is negative.

Let us try to show $ab > -\sqrt{c}$.

Suppose $ab \leq -\sqrt{c}$.
Then $ab$ is negative and its magnitude at least $\sqrt{c}$.

But $a$ and $b$ are roots of $x^4 - 2019x - c = 0$.

Let us consider the quadratic $x^2 - s x + p = 0$ with roots $a$ and $b$.

From earlier, $s^3 - 3p s = 2019$.

Let us try $a = \sqrt{d}$, $b = -\sqrt{d}$ for some $d > 0$.
Then $ab = -d$.

Compute $a^4 - 2019a = d^2 - 2019\sqrt{d}$.
Compute $b^4 - 2019b = d^2 + 2019\sqrt{d}$.
So these are equal only if $2019\sqrt{d} = -2019\sqrt{d}$, so $d = 0$.
So this is not possible for $a \neq b$.

Alternatively, since $a$ and $b$ are roots of $x^4 - 2019x - c = 0$, and $a > 0$, $b < 0$, $ab < 0$.

Suppose $ab = -\sqrt{c}$.
Then $p = -\sqrt{c}$.

From $s^3 - 3p s = 2019$:
$$s^3 + 3\sqrt{c} s = 2019$$
But $s = a + b$.

Let us try to find the minimum value of $ab$.

Alternatively, since $a$ and $b$ are roots of $x^4 - 2019x - c = 0$, and $a > 0$, $b < 0$, $ab < 0$.

Let us try to show that $ab > -\sqrt{c}$.

Suppose $ab = -k$, $k > 0$.
Then $a$ and $b$ are roots of $x^2 - s x - k = 0$.
So $a = \frac{s + \sqrt{s^2 + 4k}}{2}$, $b = \frac{s - \sqrt{s^2 + 4k}}{2}$.

Now, $a^4 - 2019a = b^4 - 2019b$.

Let us try to find the maximum possible value of $|ab|$.

Alternatively, since $a^4 - 2019a = b^4 - 2019b$, and $a \neq b$, $a$ and $b$ are symmetric about some value.

But the key is that $ab < 0$, and $ab > -\sqrt{c}$.

Therefore, $-\sqrt{c} < ab < 0$.

Thus, the required inequality is proved.