Let positive numbers $a, b, c$ satisfy the inequality $5abc > a^3 + b^3 + c^3$. Let us show that there exists a triangle with the sides $a, b, c$. On the contrary, suppose that there is no such triangle. Then for $a, b, c$ at least one of the triangle inequalities is not valid. Let, e.g. $c \ge a+b$, i.e. $c = a+b+x$, where $x \ge 0$. Then from the initial inequality it follows that
$$5ab(a + b + x) > a^3 + b^3 + (a + b)^3 + 3(a + b)^2 x + e(a + b)x^2 + x^3$$
or
$$2a^2b + 2ab^2 > 2a^3 + 2b^3 + (a+b)^3 + abx + 3(a^2+b^2)x + 3(a+b)x^2 + x^3$$
Since the last four summands at the left side are nonnegative, we have
$$\begin{aligned} 2a^2b + 2ab^2 &> 2a^3 + 2b^3 & \Leftrightarrow \\ a^3 + b^3 &- a^2b - ab^2 < 0 & \Leftrightarrow \\ (a+b)(a-b)^2 &< 0 \end{aligned}$$
which is impossible.
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In fact, assume the contrary: Assume that there are three positive reals $a, b, c$ satisfying $5abc > a^3 + b^3 + c^3$, but they are not the sidelengths of a triangle. Since everything is symmetric, we can WLOG assume that $a \ge b \ge c$. Then, $c+a > b$ (since $a \ge b$) and $a+b > c$ (since $a \ge c$), so that we must have $b+c \le a$ (else, the positive reals $a, b, c$ would be the sidelengths of a triangle). Hence, $c+a-b > 0$, $a+b-c > 0$ and $b+c-a \le 0$, so that
$$(b+c-a)(c+a-b)(a+b-c) \le 0$$
Thus,
$$5abc - (b+c-a)(c+a-b)(a+b-c) \ge 5abc$$
On the other hand, $5abc > a^3 + b^3 + c^3$. Thus,
$$\begin{aligned} 5abc - (b+c-a)(c+a-b)(a+b-c) &> a^3 + b^3 + c^3 & \Leftrightarrow \\ 10abc + (a^3 + b^3 + c^3) - (a+b+c)(bc+ca+ab) &> a^3 + b^3 + c^3 & \Leftrightarrow \\ 10abc &> (a+b+c)(bc+ca+ab) & \Leftrightarrow \end{aligned}$$
Division by $abc$ yields
$$10 > (a+b+c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$
Now, since $10 = 3^2 + 1$, the IMO 2004 problem 4 yields that $a, b, c$ are the sidelengths of a triangle, contradicting our assumption. Thus, our assumption was wrong, and we are done. ☐