60th Belarusian Mathematical Olympiad

Set $a+b+c=x$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=y$. Let $k=\frac{x}{6}$, then $$a+b-5c=6(k-c),b+c-5a=6(k-a),c+a-5b=6(k-b).$$ It follows that $$(a+b+c)\left(\frac{1}{a+b-5c}+\frac{1}{b+c-5a}+\frac{1}{c+a-5b}\right)=\frac{x}{6}\left(\frac{1}{k-a}+\frac{1}{k-b}+\frac{1}{k-c}\right).(1)$$ Adding the fractions in the right-hand side of (1), we obtain $\frac{A}{B}$, where $$A=(k-a)(k-b)+(k-b)(k-c)+(k-c)(k-a)\text{ and }B=(k-a)(k-b)(k-c).(2)$$ Let $z=abc$, then $ab+bc+ca=yz$. From (2) it follows $$A=3k^2-2zk+yz\text{and}B=k^3-z{k}^2+yz-z.(3)$$ Since $x=6k$ and $xy=\frac{27}{2}$, we have $y=\frac{9}{4k}$. Substitute for $x$ and $y$ in (3) to obtain $$A=3k^2-2\cdot 6k\cdot k+\frac{9}{4k}z=9\left(-k^2+\frac{z}{4k}\right),B=k^3-6k\cdot k^2+\frac{9}{4k}zk-z=5k\left(-k^2+\frac{z}{4k}\right).$$ Thus $$\frac{A}{B}=\frac{9}{5k},\text{and}\frac{x}{6}\cdot \frac{A}{B}=k\cdot \frac{9}{5k}=\frac{9}{5}.$$ Hence $$(a+b+c)\left(\frac{1}{a+b-5c}+\frac{1}{b+c-5a}+\frac{1}{c+a-5b}\right)=\frac{9}{5}.$$