Outline from Contest: This solution is quite creative, you do not need to be this creative. We replace $1$ by $(a+b+c)^2$ and let $S$ be the sum. We claim that $S = 3$ is the minimum which is achieved when $(a,b,c)$ is a permutation of $(-3,1,1)$. It is then possible to see that if we define $x = a+b, y = b+c, z = c+a$, then $$S = \sum_{cyc} \frac{xy}{z}-3$$Then, notice that two of $x,y$ is negative and the last positive, so let $x = p, y = q$; the crux is the following claim: $\textbf{Lemma:}$ $pqz \geq 8$. For this, notice that $p+q = z+2$ and that $abc \leq -3$ implies that $(p-1)(q-1)(p+q-1) \geq 3$, then it is possible to show that $p, q > 1$ and replace $p \to \alpha+1$ and $q \to \beta +1$ and use $(\frac{\alpha+\beta}{2})^2 \geq \alpha \cdot \beta$ to show that $(\alpha+1)(\beta+1)(\alpha+\beta) \geq 8$ which proves the Lemma. Then it is easy to finish. $S = \sum_{cyc} \frac{xy}{z} -3 = \sum_{cyc} \frac{pq}{z}-3 \geq 3 \cdot \sqrt[3]{pqz}-3 \geq 3 \cdot 2 - 3 = 3$, we are done. Remark: Problems are ordered by difficulty (mod 4), only $5$ of the problems were released due to confidentiality.

Let $a,b,c$ be reals such that $a+b+c =1$ and $abc\geq 3.$ Prove that$$\frac{ab+1}{a+b}+\frac{bc+1}{b+c}+\frac{ca+1}{c+a}\leq -3$$$$\frac{ab+c}{a+b}+\frac{bc+a}{b+c}+\frac{ca+b}{c+a}\leq -6$$

Let $a,b,c$ be reals such that $a+b+c =1$ and $abc\geq 3.$ Prove that$$\frac{ab+k}{a+b}+\frac{bc+k}{b+c}+\frac{ca+k}{c+a}\leq \frac{k-7}{2}$$$$\frac{ab+kc}{a+b}+\frac{bc+ka}{b+c}+\frac{ca+kb}{c+a}\leq - \frac{5k+7}{2}$$Where $k>0.$

Rewrite $\frac{ab+1}{a+b}=\frac{ab+(a+b+c)^2}{a+b}=\frac{(a+c)(b+c)}{a+b}+a+b+c=\frac{(a+c)(b+c)}{b+c}-1$. We thus wish to find the minimum of $\frac{(a+c)(b+c)}{a+b}+\frac{(b+a)(c+a)}{b+c}+\frac{(c+b)(a+b)}{c+a}$. Notice when $(a,b,c)=(1,1,-3)$ (to get $6$) we will have the three fractions equal, which means AMGM is possible: it suffices to show $(a+b)(b+c)(c+a)\ge8$. Rewrite as $8\le (a+b+c)(ab+bc+ca)-abc=-ab-bc-ca-abc$, suffices to show $-ab-bc-ca\ge 5$. WLOG $a,b$ positive, $c$ negative. Then $ab(a+b+1)\ge 3\implies b(a+b+1)\ge \frac{3}{a}$. Now $-ab-bc-ca=(a+b)(a+b+1)-ab=b(a+b+1)+a^2+a\ge a^2+a+\frac{3}{a}$. It suffices to show this is $\ge5$, which is true as it factorises to $(a-1)^2(a+3)\ge0$.

@above Good idea, but you have to be careful about applying your AM-GM, since some of the fractions may be negative.

L567 wrote: Determine the smallest possible value of the expression $$\frac{ab+1}{a+b}+\frac{bc+1}{b+c}+\frac{ca+1}{c+a}$$where $a,b,c \in \mathbb{R}$ satisfy $a+b+c = -1$ and $abc \leqslant -3$ $$\prod_{cyc}(a+b)=(a+b+c)(ab+ac+bc)-abc\geq3-(ab+ac+bc)=\sum_{cyc}(3a^2+5ab)=\frac{5}{2}+\frac{1}{2}(a^2+b^2+c^2)>0.$$Thus, it's enough to prove a fourth degree inequality, for which by $uvw$ it's enough to check an equality case of two variables.