India_2017

Let us consider a cubic polynomial whose roots are $a$, $b$, $c$. We get $P(x)=x^3-p{x}^2+qx-r$, where $p=a+b+c$, $q=ab+bc+ca$ and $r=abc$. We observe that $$\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-c)(b-a)}+\frac{c^n}{(c-a)(c-b)}=\sum _{\text{cyclic}}\frac{-a^n(b-c)}{(a-b)(b-c)(c-a)}$$ Let us write $$S_n=\sum _{\text{cyclic}}\frac{a^n(b-c)}{-(a-b)(b-c)(c-a)}$$ It is easy to see that $S_1=0$ and $S_2=1$. Since $a$, $b$, $c$ are the roots of $P(x)=0$, we have $a^3-p{a}^2+qa-r=0$, $b^3-p{b}^2+qb-r=0$, $c^3-p{c}^2+qc-r=0$. Multiply the first by $b-c$, the second by $c-a$ and the third by $a-b$, and adding all these and dividing the sum by $-(a-b)(b-c)(c-a)$, we obtain $S_3-p{S}_2+q{S}_1=0$. Hence $S_3=p$. Now multiply the first by $a$, the second by $b$ and the third by $c$ and divide throughout by $-(a-b)(b-c)(c-a)$ to get $S_4-p{S}_3+q{S}_2-r{S}_1=0$. Hence $S_4=p^2-q$. Similarly, $S_5=p(p^2-q)-qp+r=p^3-2pq+r$. We also get $S_6-p{S}_5+q{S}_4-r{S}_3=0$. This gives $$S_6=p(p^3-2pq+r)-q(p^2-q)+rp=p^4-3p^{2}q+2pr+q^2.$$ We can write it as $S_6=p^2(p^2-3q)+2pr+q^2$. But $p^2-3q=(a+b+c{)}^2-3(ab+bc+ca)=a^2+b^2+c^2-ab-bc-ca>0$. Hence $$S_6>2pr+q^2=2abc(a+b+c)+(ab+bc+ca{)}^2\ge 6+9=15.$$