Callous Constraint

Relevant wiki: Classical Inequalities - Problem Solving - Advanced

Since the constraint $a^3 + b^3 + c^3 = \big(3 + \tfrac{1}{666}\big)abc$ and the target function $\tfrac{a}{b} + \tfrac{b}{c} + \tfrac{c}{a}$ are both homogeneous (of degrees $3$ and $0$ respectively), we can apply the additional condition $a+b+c = 3$ without loss of generality.

Suppose now that $a,b,c$ are positive reals such that $a^3 + b^3 + c^3 \; = \; (3 + 27\alpha)u \qquad \qquad a + b + c \; = \; 3 \qquad \qquad abc \; = \; u$ for positive constants $\alpha,u$ . Then $(3 + 27\alpha)u \; = \; (a+b+c)^3 - 3(a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2) - 6abc \; = \; 27 - 3(a+b+c)(ab+ac+bc) + 3abc$ so that $ab + ac + bc \; = \; 3(1 - \alpha u)$ Since $a,b,c$ are positive this means that $0 < \alpha u < 1$ . Now $a,b,c$ are the roots of the cubic equation $0 \; = \; f(X) \; = \; X^3 - 3X^2 + 3(1 - \alpha u)X - u \; = \; (X - 1)^3 - 3\alpha u(X - 1) + 1 - (1 + 3\alpha)u$ Since $f(X)$ has three positive real roots, its derivative has two positive real roots. Since $f'(X) \; = \; 3\big[(X - 1)^2 - \alpha u\big]$ we see that the turning points of $f$ occur at $\lambda_\pm = 1 \pm \sqrt{\alpha u}$ , which are indeed both positive, since $0 < \alpha u < 1$ . Note that $f(x) < 0$ for all $x\le 0$ , and hence that all real roots of $f(X)$ are positive. Since $f(\lambda_\pm) \; = \; \mp2\alpha u \sqrt{\alpha u} + 1 - (1 + 3\alpha)u$ and, in order that $f(X)$ have three positive real roots, we require that $f(\lambda_-) \ge 0 \ge f(\lambda_+)$ , it follows that $\big|(1 + 3\alpha)u - 1\big| \; \le \; 2\alpha u \sqrt{\alpha u}$ In other words, we require that $4\alpha^3 u^3 - \big((1 + 3\alpha)u-1\big)^2 \; \ge \; 0$ In our case of $\alpha = \tfrac{1}{27\times666}$ , numerical calculations show that we must have $0.99983236525 \le u \le 0.99983402340$ .

Consider the two quantities $p \; = \; \tfrac{a}{b} + \tfrac{b}{c} + \tfrac{c}{a} \qquad \qquad q \; = \; \tfrac{a}{c} + \tfrac{c}{b} + \tfrac{b}{a}$ noting that the difference between $p$ and $q$ is simply a matter of the labelling of the roots of $f(X)$ . Playing with Vieta's formulae, we can show that $p+q \; = \; \tfrac{9}{u} - 3(3\alpha+1) \qquad \qquad pq \; = \; 9 + 54\alpha - \tfrac{27}{u} + \tfrac{27(1-\alpha u)^3}{u^2}$ and so $p,q$ are the roots of the quadratic equation $h(X) \; = \; X^2 - \big( \tfrac{9}{u} - 3(3\alpha+1) \big)X + 9 + 54\alpha - \tfrac{27}{u} + \tfrac{27(1-\alpha u)^3}{u^2} \; = \; 0$ Thus, for any particular values of $\alpha,u$ , the smallest possible value of $\tfrac{a}{b} + \tfrac{b}{c} + \tfrac{c}{a}$ is the smaller of the two zeros of $h(X)$ , call it $L_\alpha(u)$ . The explicit formula for $L_\alpha(u)$ is $L_\alpha(u) \; = \; \tfrac{3}{2u}\left[3 - (3\alpha+1)u - \sqrt{3\left[4\alpha^3u^3 - ((3\alpha+1)u-1)^2\right]}\right]$ which is reassuring, since the term inside the square root only gives a real value for $L_\alpha(u)$ when $4\alpha^3u^3 \ge ((3\alpha+1)u-1)^2$ , which is the condition we have already determined as being necessary.

In our particular case of $\alpha = \tfrac{1}{27 \times 666}$ , we want to minimize $L_\alpha(u)$ over the interval $0.99983236525 \le u \le 0.99983402340$ . Numerical calculations (solving for the exact value of $u$ where $L_\alpha(u)$ has a turning point requires solving an order $6$ polynomial) show that this occurs at $u = 0.99983391233$ , taking the least value $3.00049619108 = 3 + \tfrac{1}{2015.35}$ . Thus we have shown that if $a,b,c$ are positive real numbers such that $a^3 + b^3 + c^3 = \big(3 + \tfrac{1}{666}\big)abc$ , then $\tfrac{a}{b} + \tfrac{b}{c} + \tfrac{c}{a} \; \ge \; 3 + \tfrac{1}{2015.35}$ Thus the smallest possible integer $k$ is $\boxed{2016}$ .

I really hope that someone can come up with a neater solution than this!