Proposed by me

A proof for $ab<0$: It is obvious that $a$ and $b$ are non-zero. Note that the condition is equivalent to $(a^4-b^4=2019(a-b)$, i.e. $(a+b)(a^2+b^2)=2019$ This means that the case $a<0$, $b<0$ is impossible. Now (WLOG) let us suppose for the sake of contradiction that $a>b>0$ We notice that since $c$ is positive from the condition one has $a^4>2019a$ and $b^4>2019b$ i.e. $a^3>b^3>2019$. Considering the function $f(x)=x^4-2019x$, we see that its derivative is $f'(x)=4x^3-2019>0$ for $x^3>2019$, so the function is strictly increasing, i.e.the equality fails to hold, contradiction. So $a$ and $b$ have differnt signs ($ab<0$)

The given equation is equivalent to $(a+b)(a^2 + b^2) = 2019$. Therefore, $a + b > 0$. We divide the solution into two parts: Part 01. $ab < 0$ Proof. We have \[ a^4 + b^4 - 2019(a+b) > 0 \]\[ a^4 + b^4 - (a^2 + b^2)(a+b)^2 > 0 \]\[ 2ab(a^2 + b^2 + ab) < 0 \]Since $a^2 + b^2 + ab > 0$, it suffices that $ab < 0$. Part 02. $-\sqrt{c} < ab$ Proof. Since $ab < 0$, it is sufficient to prove that $c > (ab)^2$. But, \[ 2c = (a^4 - 2019a) + (b^4 - 2019b) = (a^4 + b^4) - 2019(a+b) > 2(ab)^2 \]which follows by $(a^2 - b^2)^2 - 2019(a+b) > 0$, that is, \[ (a^2 - b^2)^2 - (a+b)^2 (a^2 + b^2) > 0 \]\[ (a^4 + b^4 - 2a^2 b^2) - (a^4 + b^4 + 2a^2 b^2 + 2a^3 b + 2ab^3) > 0 \]\[ 2ab (a+b)^2 < 0 \]

A proof for $ab>-\sqrt{c}$ can also be given in this way: Just note that it suffices to prove that $c^2>a^4b^4 \Longleftrightarrow a^4b^4-2019ab(a^3+b^3)+2019a^2b^2>0 \Longleftrightarrow a^2b^2>ab(a^3+b^3)$. Now $a^3+b^3=(a+b)(a^2+b^2-ab)>0$ since $a+b>0$ and $ab<0$ so the inequality is obvious and the proof is completed

For the first one: We have that $(a+b)(a^2+b^2)=2019$. What is more, $c^{2}=(a^4-2019a)(b^4-2019)=...=ab[(ab)^3-(a+b)^2(a^2+b^2)(a^2+ab+b^2)-(a+b)^2(a^2+b^2)]$. It is obvious that $(ab)^3-(a+b)^2(a^2+b^2)(a^2+ab+b^2)-(a+b)^2(a^2+b^2)<0$ while $c^2>0$. So $ab<0$.

Consider the polynomial $P(x)=x^4-2019x-c$. Then $P(x)$ has at least two real roots, $x=a$ και $x=b$, and so, we can write $P(x)=(x^2-(a+b)x+ab)(x^2+px+q),$ for some $p,q\in \mathbb{R}$. Since the coefficient of $x^3$ in $P(x)$ is equal to $0$, we see that $p=a+b$, and so $P(x)=(x^2-(a+b)x+ab)(x^2+(a+b)x+q).$ Applying the distributive property on the right-hand side, and equating the coefficients of the corresponding terms, we have $\begin{aligned} q&=a^2+ab+b^2,\\ ab^2+ab^2-(a+b)q&=-2019,\\ abq&=-c.\\ \end{aligned}$ Hence $q>0$, which combined with $c>0$ implies that $\boxed{ab=-c/q<0}$. Moreover, we have $2019=(a^4-b^4)/(a-b)=(a+b)(a^2+b^2)$. Therefore, $a+b>0$ and so $\begin{aligned} (a+b)^2>0&\iff a^2+ab+b^2>-ab\\ &\iff q>-ab\\ &\iff -q<ab\\ &\iff c=-qab>(ab)^2\\ &\iff \sqrt{c}>\sqrt{(ab)^2}=|ab|=-ab\\ &\iff \boxed{-\sqrt{c}<ab}.\\ \end{aligned}$

Only Vieta is needed. The function $f(x)=x^4-2019x-c$ in increasing and then decreasing, so $f$ has exactly two positive roots, $a$ and $b$. This means that the other two are complex and conjugate, let $r$ and $\overline{r}$. From Vieta, we have 1) $|r|^2ab=-c$ (1), so $ab<0$. 2) $r+\overline{r}=-a-b$ and 3) $\displaystyle{(a+b)(r+\overline{r})=\frac{c-(ab)^2}{ab}}$, so $-(a+b)^2=\frac{c-(ab)^2}{ab}$, but $ab<0$, so $c>(ab)^2$, which is the desired.

falantrng wrote: Proposed by me You are from Saudi Arabia.

Similar problem, see problem 3, from 5/27/2019: https://vnexpress.net/giao-duc/loi-giai-de-toan-chuyen-truong-pho-thong-nang-khieu-3929752.html?fbclid=IwAR2kLM2RJyZhLvjpzBhXalRPLFuJxpCBjcL58c6plBdqhSk7UdVsQkfvaoA

Lukaluce wrote: Let $a$, $b$ be two distinct real numbers and let $c$ be a positive real numbers such that $a^4 - 2019a = b^4 - 2019b = c$. Prove that $- \sqrt{c} < ab < 0$. Consider the polynomial $x^4 - 2019x -c.$ Let $r_1, r_2, r_3, r_4$ be its roots. By Vieta's, we obtain $\sum_{i=1}^4 r_i = \sum_{1 \leq i < j \leq 4} r_i r_j = 0,$ and so $\sum_{i=1}^4 r_i ^2 = 0.$ Thus, not all of the roots can be real (or they will be zero, which is not a root.) The problem tells us there are real roots, and so it must be that two roots, $r_1, r_2$ are real while $r_3, r_4$ are nonreal, complex conjugates. To show $ab < 0,$ we can look at behavior of $x^4 - 2019x,$ because our polynomial is just a translation down of this. $x^4 - 2019x$ has roots zero and $(2019)^{1/3},$ (and these are the only real roots!), so by taking derivatives, we see this polynomial is always decreasing on $x<0,$ and always increasing on $x>(2019)^{1/3}.$ Then, $x^4 - 2019x - c$ must have a root smaller then zero and a root larger then $(2019)^{1/3},$ and one so one of $a,b,$ is positive, the other negative: Their product is less then zero. Let $r_1 = a, r_2 = b.$ Then, $-\sqrt{c} < ab \Longleftrightarrow c > (r_1 r_2)^2,$ or $-r_1 r_2 r_3 r_4 > r_1 ^2 r_2^2,$ by vieta's again. Dividing out $-r_1 r_2$ on both sides, we see it is equivalent to showing $r_3 r_4 + r_1 r_2 > 0.$ Recall that $\sum_{1 \leq i < j \leq 4} r_i r_j = 0.$ Note that $r_1 r_3 + r_1 r_4 + r_2 r_3 + r_2 r_4 = (r_1 + r_2)(r_3 + r_4) < 0,$ because $\sum_{i=1}^4 r_i = 0,$ and so $r_1 + r_2, r_3 + r_4$ can't both be positive or negative. Additionally, neither is zero, because we can check that if $p$ is a solution, than $q$ is not a solution.Thus, $r_3 r_4 + r_1 r_2 = \sum_{1 \leq i < j \leq 4} r_i r_j - (r_1 + r_2)(r_3 + r_4) = 0 - (r_1 + r_2)(r_3 + r_4) > 0,$ and we are done.

DievilOnlyM wrote: Similar problem, see problem 3, from 5/27/2019: https://vnexpress.net/giao-duc/loi-giai-de-toan-chuyen-truong-pho-thong-nang-khieu-3929752.html?fbclid=IwAR2kLM2RJyZhLvjpzBhXalRPLFuJxpCBjcL58c6plBdqhSk7UdVsQkfvaoA

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Lukaluce wrote: Let $a$, $b$ be two distinct real numbers and let $c$ be a positive real numbers such that $a^4 - 2019a = b^4 - 2019b = c$. Prove that $- \sqrt{c} < ab < 0$. $\left\{\begin{array}{ll}a^4=&2019a+c\\b^4=&2019b+c\end{array}\right.\overset{-}{\Longrightarrow}(a-b)(a+b)(a^2+b^2)=2019(a-b)\Longrightarrow a+b>0$ $\left\{\begin{array}{ll}a^3=&2019+\frac{c}{a}\\b^3=&2019+\frac{c}{b}\end{array}\right.\overset{-}{\Longrightarrow}(a-b)(a^2+ab+b^2)=-\frac{c(a-b)}{ab}\Longrightarrow ab<0$ $\left\{\begin{array}{ll}a^2=&\frac{2019}{a}+\frac{c}{a^2}\\b^2=&\frac{2019}{b}+\frac{c}{b^2}\end{array}\right.\overset{+}{\Longrightarrow}a^2+b^2=\frac{2019(a+b)}{ab}+\frac{c(a^2+b^2)}{a^2b^2}\Longrightarrow(a^2+b^2)(a^2b^2-c)=2019ab(a+b)\Longrightarrow a^2b^2-c\le0\Longrightarrow ab>-\sqrt{c}$

Since $c>0$, we have $ab\neq 0$, thus $2019=\frac{a^4-c}{a}=\frac{b^4-c}{b}\cdots (E_1)$ By the symmetry of $a,\ b$, W.L.O.G. $a<b$. $(E_1)$ shows that three points $A(a,\ a^4),\ K(0,\ c), B(b,\ b^4)$ are colinear and points $A,\ B$ are on the graph of $y=x^4$, and these points lie on the same line $y=2019x+c$. Therefore, one of the points $A,\ B$ should be located on the left side and the other on the right side of the $y$ axis, respectively, or $a<0<b$.Now, from $(E_1)$, we get $c(a-b)=-ab(a^3-b^3).$ From $a\neq b$, yielding $c=-ab(a^2+ab+b^2) \cdots(E_2)$. From $(E_2)$, we have $-\sqrt{c}<ab\Longleftrightarrow \sqrt{c}>-ab>0$ $\Longleftrightarrow c>(-ab)^2>0 \Longleftrightarrow -ab(a^2+ab+b^2)>(-ab)^2>0$, dividing both sides by $-ab>0$, yielding $(a+b)^2>0$, which is true.Because $a,\ b$ are distinct real numbers and $ab<0$, $(a+b)^2$ can't be zero. Because if $a+b=0$, then by the given condition, we have $a^4-2019a=(-a)^4-2019(-a)\Longleftrightarrow a=0$ which contradicts $a\neq 0.$ Comment 1 : For proof of $ab<0$, $a^{2n} - 2019a = b^{2n} - 2019b = c\ (a\neq b, c>0,\ n=1,\ 2,\ \cdots)$ is also interesting ! I wonder if for proof of $ab<0$, $a^{6} - 2019a = b^{6} - 2019b = c\ (a\neq b, c>0,\ n=1,\ 2,\ \cdots)$ was more intesting, and impressive ^^ Comment 2 : Rewrite the given condition as $a^4 - 2019a = c \Longleftrightarrow a^4 = 2019a + c$ and $b^4 = 2019b + c$ , which shows a, b are the abscissae of the intersections of the curve $y = x^4$ and the line $y = 2019 x+c$. Since $c>0$, one of $a, b$ is positive and the other is negative, yielding $ab<0$.

It's easy to see that $f(x)=x^4-2019x$ is positive iff $x \in (- \infty, 0)$ or $x \in (\sqrt[3]{2019}, \infty)$, and that $f(x)$ is strictly decreasing and increasing on these intervals respectively. Thus we must have $ab<0$. Now using the fact that $(a-b)^2>a^2+b^2$, we get $$a^4+b^4-2a^2b^2 = (a+b)^2(a-b)^2>(a+b)^2(a^2+b^2)=2019(a+b)$$$$\Rightarrow c = \frac{a^4+b^4-2019(a+b)}{2} > a^2b^2$$as desired.

We prove the bounds separately. We show $ab<0$. Notice that $a^4-2019a=b^4-2019b\implies (a+b)(a^2+b^2)=2019$($a$ and $b$ distinct). Notice that $a$ and $b$ cannot both be negative as then the RHS of the previous expression would have had to be negative. $a$ and $b$ also cannot both be positive because if there were, then $a,b\geq (2019^{\frac{1}{3}},$ so $(a+B)(a^2+b^2)\geq 4\cdot 2019,$ contradiction. Therefore, one of them is positive and one of them is negative. We show $-\sqrt{c}<ab$. Notice that because both sides are negative, this statement is equivalent to $c>a^2b^2$. We will prove this equivalent statement. Summing the two equations, we get that $a^4+b^4-2019(a+b)=2c\implies 2c\geq 2a^2b^2-2019(a+b)$. Notice that we want a minimum bound on $a+b,$ so AM-GM will not help in the direct way, but we can use $(a+b)(a^2+b^2)=2019$ and AM-GM together. We get that $a^2+b^2\geq 2ab,$ and since $a+b$ is positive(because 2019 is positive), we know that $a+b\leq \frac{2019}{2ab}$. This is already strong enough, as $2c\geq 2a^2b^2-\frac{2019^2}{2ab}\implies c\geq a^2b^2-\frac{2019^2}{4ab}$. Because $ab$ is negative, $\frac{2019^2}{4ab}<0,$ so we are done.

Datis-Kalali wrote: A proof for $ab<0$: It is obvious that $a$ and $b$ are non-zero. Note that the condition is equivalent to $(a^4-b^4=2019(a-b)$, i.e. $(a+b)(a^2+b^2)=2019$ This means that the case $a<0$, $b<0$ is impossible. Now (WLOG) let us suppose for the sake of contradiction that $a>b>0$ We notice that since $c$ is positive from the condition one has $a^4>2019a$ and $b^4>2019b$ i.e. $a^3>b^3>2019$. Considering the function $f(x)=x^4-2019x$, we see that its derivative is $f'(x)=4x^3-2019>0$ for $x^3>2019$, so the function is strictly increasing, i.e.the equality fails to hold, contradiction. So $a$ and $b$ have differnt signs ($ab<0$) For the part for getting a contradictoin from a>b>0 is easier with Descartes Sign Rule

Let $s=a+b$, $p=ab$. By dividing with $a-b \neq 0$ in $a^4 - b^4 = 2019(a-b)$ we get $a^3 + a^2b + ab^2 + b^3 = 2019 \Leftrightarrow s(s^2-2p) = 2019$. Hence $s>0$ and so without loss of generality $a>0$. From $c>0$ we now obtain $a^3 > 2019$. Suppose, for contradiction, that $ab\geq 0$. Since $b\neq 0$ (otherwise $c=0$), we have $b>0$ and so $b^3 > 2019$. Thus necessarily $2s^3 - 4sp = 4038 < a^3 + b^3 = s^3 - 3sp$, giving $s(s^2-p) < 0$ - the latter contradicting $s>0$ and $s^2 \geq 4p$. Hence $ab < 0$. Suppose, for contradiction, that $-\sqrt{c} \geq ab$. Then $2c \leq 2p^2$. On the other hand, $2c = a^4 + b^4 - 2019(a+b) = s^4 - 4s^2p + 2p^2 - 2019s$. Thus necessarily $s^4 - 4s^2p - 2019s \leq 0$ and now $s(s^2-2p) = 2019$ gives $-2s^2p \leq 0$, contradicting the already proven $p=ab<0$.

Datis-Kalali wrote: A proof for $ab<0$: It is obvious that $a$ and $b$ are non-zero. Note that the condition is equivalent to $(a^4-b^4=2019(a-b)$, i.e. $(a+b)(a^2+b^2)=2019$ This means that the case $a<0$, $b<0$ is impossible. Now (WLOG) let us suppose for the sake of contradiction that $a>b>0$ We notice that since $c$ is positive from the condition one has $a^4>2019a$ and $b^4>2019b$ i.e. $a^3>b^3>2019$. Considering the function $f(x)=x^4-2019x$, we see that its derivative is $f'(x)=4x^3-2019>0$ for $x^3>2019$, so the function is strictly increasing, i.e.the equality fails to hold, contradiction. So $a$ and $b$ have differnt signs ($ab<0$) ab is infact -332

First let's prove $0>ab$. $(a^2+b^2)(a+b)(a-b)=a^4-b^4=2019(a-b)$ $a,b$ are distinct $\implies (a^2+b^2)(a+b)=2019$ Note that $a+b>0$. If $a>0$ then $a^3>2019=a^3+(a^2b+ab^2+b^3) \implies 0>b$ If $a<0$ then $a^3+(a^2b+ab^2+b^3)=2019>a^3$ $\implies b(a^2+ab+b^2)>0$ So $b>0$ $\implies ab<0$ Secondly let's prove that $-\sqrt{c}<ab$. Let $a=x$,$b=-y$ such that $x,y>0$. $-xy>-\sqrt{x^4-2019x}$ $xy<\sqrt{x^4-2019x}$ Both sides are positive then $x^2y^2<x^4-2019x$ $xy^2<x^3-2019=y^3-xy^2+x^2y$ $2xy^2<y(x^2+y^2)$ $2xy<x^2+y^2$ is obviously true.

Solution for $-\sqrt{c} < ab$ with information that $ab < 0$: Without loss of generality, assume from $ab < 0$ that $a > 0 > b$. Let $x = -b$. Now we have a condition $a^4 - 2019a = x^4 + 2019x = c$ and we want to show that $a^4x^4 < c^2$, i.e. $$a^4x^4 < (a^4 - 2019a)(x^4 + 2019x) = a^4x^4 - 2019ax^4 + 2019a^4x - 2019^2ax$$Which is equivalent to: $$x^3 + 2019 < a^3$$Now suppose for the sake of contradiction that $x^3 + 2019 \ge a^3$. Now: $$xa^3 \le x(x^3 + 2019) = x^4 + 2019x = a^4 - 2019a = a(a^3 - 2019) \le ax^3$$Which is equivalent to $a \le x$ which isn't true because $a^4 - 2019a = x^4 + 2019x$, hence we have a contradiction and the result follows.