Well... Obviously the answer is not 5997

Start by considering the function $f_k(a) \; = \; a^k + (3-a)^k - a(3-a) \hspace{2cm} 0 \le a \le 3$ for any $k > 0$ .

If $k > 1$ then, for any $0 < a < 3$ , $f_k'(a) \,=\, k(a^{k-1} - (3-a)^{k-1}) + a - (3-a) \; = \; (2a-3)\big[k(k-1)u^{k-2} + 1\big]$ for some $u$ between $a$ and $3-a$ (using the Mean Value Theorem). Thus we deduce that $f_k'(a) < 0$ for $0 < a < \tfrac32$ , while $f_k'(a) > 0$ for $\tfrac32 < a < 3$ , and hence $f_k(a)$ is minimized at $a=\tfrac32$ . Additionally, if $k=1$ then $f_k(a) = 3 - a(3-a)$ is also minimized at $a = \tfrac32$ .

If $0 < k < 1$ then since $\begin{aligned} f_k'(a) & = \; k\big[a^{k-1} - (3-a)^{k-1}\big] + 2a - 3 \\ f_k''(a) & = \; k(k-1)\big[a^{k-2} + (3-a)^{k-2}\big] + 2 \\ f_k'''(a) & = \; k(k-1)(k-2)\big[a^{k-3} - (3-a)^{k-3}\big] \end{aligned}$ we see that $f_k'''(a)$ has a unique zero at $a=\tfrac32$ , and hence that $f_k''(a)$ has a global maximum at $\tfrac32$ . Since $\big(\tfrac23\big)^{2-k}f_k''(\tfrac32) \; = \; 2\big[\big(\tfrac23\big)^{2-k} + k(k-1)\big] \; \ge\; 2\big[\big(\tfrac23\big)^2 - \tfrac14\big] \; > \; 0$ we deduce that this global maximum is positive. Thus we can obtain the following sketches for the graphs of $f_k''',f_k'',f_k',f_k$ , from which it is clear that $f_k$ achieves its global minimum at $a = \tfrac32$ .

Thus $f_k(a) \; \ge \; f_k(\tfrac32) \; = \; 2\big(\tfrac32\big)^k - \tfrac94 \hspace{2cm} 0 \le a \le 3 \;,\; k > 0$ and so we see that $f_k(a) \ge 0$ for all $0 \le a \le 3$ provided that $k \ge \kappa = \log_{1.5}1.125$ .

Now consider the general function $F_k(a,b,c) \; = \;a^k + b^k + c^k - ab - ac - bc$ subject to the constraints $a,b,c \ge 0$ and $a+b+c = 3$ . A minimum of $F_k$ on the boundary of this domain corresponds to a minimum of $f_k$ , and hence is nonnegative precisely when $k \ge \kappa$ . We need to consider the extremal points of $F_k$ at interior points of the constraint domain. At such a point, the method of Lagrange multipliers tells us that there must exist $\lambda$ such that $\left(\begin{array}{c} ka^{k-1} - b - c \\ kb^{k-1} - a - c \\ kc^{k-1} - a - b \end{array}\right) \; = \; \lambda\left(\begin{array}{c} 1 \\ 1 \\ 1\end{array} \right)$ and hence that $g_k(a) \; = \; g_k(b) \; = \; g_k(c) \; = \; \lambda + 3$ where $g_k(x) = kx^{k-1} + x$ for $0 < x < 3$ .

If $k \ge 1$ then $g_k'(x) = k(k-1)x^{k-2} + 1 > 0$ for all $0 < x < 3$ , and hence $g_k$ is an increasing function of $x$ , Thus we deduce that $a = b = c = 1$ , and $F_k(1,1,1) = 0$ .

If $0 < k < 1$ then we see that $g_k'(x) = 0$ at the unique point $\lambda_k = \big[k(1-k)\big]^{\frac{1}{2-k}}$ , noting that $0 < \lambda_k < 1$ . Moreover $g_k$ is strictly decreasing for $x < \lambda_k$ , and strictly increasing for $x > \lambda_k$ . Thus, for any suitable $\mu$ , there are at most $2$ distinct points $a$ between $0$ and $3$ for which $g_k(a) = \mu$ . Thus we have three cases to consider:

• We have $a = b = c = 1$ , in which case $F_k(a,b,c) = 0$ .
• We have $a = b < c = 3-2a$ , where $g_k(a) = g_k(3-2a)$ and $0 < a < 1$ . Numerical calculations show that there is a unique such value $a$ for each $0 < k < 1$ , and the graph of $F_k(a,a,3-2a)$ for these values of $a$ show that the values at these extremal points are all positive.
• We have $a = b > c = 3-2a$ , where $g_k(a) = g_k(3-2a)$ and $1 < a < \tfrac32$ . Numerical calculations show that there is a unique such value $a$ for each $0 < k < 1$ , and the graph of $F_k(a,a,3-2a)$ for these values of $a$ show that the values of these extremal points are all positive.

Thus we deduce that, for any $0 < k < 1$ , the only extrema of $F_k$ , subject to the constraints $a,b,c > 0$ and $a+b+c = 3$ , are non-negative.

In summary, then, the inequality $a^k + b^k + c^k \; \ge\; ab + bc + ac$ holds for all $a,b,c \ge 0$ such that $a + b + c = 3$ precisely when $k \ge \kappa = \log_{1.5}1.125$ . This makes the desired answer $\lfloor 1999 \kappa \rfloor = \boxed{580}$ .