60th Belarusian Mathematical Olympiad

In view of $a+b+c=0$, we rewrite the left hand side of the required inequality in the form $$\begin{array}{rl}M& =\frac{a^3(a+c)(a+b)}{(a-c)(a-b)}+\frac{b^3(b+a)(b+c)}{(b-a)(b-c)}+\frac{c^3(c+b)(c+a)}{(c-b)(c-a)}=\\ & =\frac{a^3(-b)(-c)}{(a-c)(a-b)}+\frac{b^3(-c)(-a)}{(b-a)(b-c)}+\frac{c^3(-a)(-b)}{(c-b)(c-a)}=\\ & =abc\left(\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-b)(c-a)}\right)=abcL.\end{array}$$ We have $$\begin{array}{rl}L& =\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-b)(c-a)}=\\ & =\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-c)(b-a)(c-b)}=\frac{N}{(a-c)(b-a)(c-b)}.\end{array}$$ Further, $$\begin{array}{rl}N& =a^2(b-c)+b^2(c-a)+c^2(a-b)=a^2(b-c)+b^{2}c-b^{2}a+c^{2}a-c^{2}b=\\ & =a^2(b-c)+(b^{2}c-c^{2}b)-(b^{2}a-c^{2}a)=a^2(b-c)+bc(b-c)-a(b+c)(b-c)=\end{array}$$ $$=(b-c)(a^2+bc-ab-ac)=(b-c)(a(a-b)-c(a-b))=(b-c)(a-b)(a-c)=(a-c)(b-a)(c-b).$$ So $L=1$, hence $M=abc$, as required.