Is It Really Balancing The Coefficients?

Relevant wiki: Classical Inequalities - Problem Solving - Advanced

Since the target function $\tfrac{a}{b}+\tfrac{b}{c} + \tfrac{c}{a}$ and the constraint equation are both homogeneous, we can apply the additional constraint $a+b+c=4$ without loss of generality. Thus we must have $\frac{ab + ac + bc}{abc} \; = \; a^{-1} + b^{-1} + c^{-1} \; = \; 4$ and hence $a,b,c$ are the roots of a cubic polynomial $f_\gamma(X) \; =\; X^3 - 4X^2 + 4\gamma X - \gamma \; = \; 0$ for some $\gamma > 0$ ; of course, $\gamma = abc$ . If we define $U \; = \; \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \qquad \qquad V \; = \; \frac{a}{c} + \frac{c}{b} + \frac{b}{a}$ then some elementary calculations with Vieta's and Newton's formulae tell us that $U + V \; = \; 13 \qquad \qquad UV \; = \; 64\gamma - 87 + 64\gamma^{-1}$ and hence $U,V$ are the roots of the quadratic equation $X^2 - 13X + UV \; = \; 0 \;,$ so that $U,V \; = \; \tfrac12\big[13 \pm \sqrt{169 - 4UV}\big] \;.$ Thus the maximum and minimum possible values of $U$ and $V$ are both achieved when $UV$ is minimized.

Over all values of $\gamma > 0$ , $UV$ is minimized when $\gamma=1$ , and the corresponding minimum value is $41$ . We just need to check that $f_1(X)$ is a polynomial with positive roots. But $f_1(X) \; = \; X^3 - 4X^2 + 4X - 1 \; = \; (X - 1)(X^2 - 3X + 1)\;,$ which does indeed have three positive roots. Thus we deduce that the maximum and minimum possible values of $U$ are the two roots of $X^2 - 13X + 41 \,=\, 0$ , and hence $m \; = \; \tfrac12\big[13 - \sqrt{5}\big] \qquad \qquad M \; = \; \tfrac12\big[13 + \sqrt{5}\big] \;.$ Note that the difference between $U$ and $V$ is just a matter of the labelling of $a,b,c$ ; if $U = m$ for $(a,b,c) = (u,v,w)$ , then $V=M$ for $(a,b,c) = (u,v,w)$ , and so $U = M$ for $(a,b,c) = (u,w,v)$ , after making a noncyclic permutation of the values of $a,b,c$ .

Thus the answer is $\left\lfloor 100 \times \tfrac12(39 + \sqrt{5})\right\rfloor \,=\, \boxed{2061}$ .