We can find easily $$S=\sum_{cyc} \frac{a^6}{(a-b)(a-c)}= p^2(p^2-3q)+2pr+q^2,$$where $p=a+b+c,q=ab+bc+ca,r=abc,$ and $p^2\ge 3q.$ Then $S=p^2(p^2-3q)+2pr+q^2> 2pr+q^2=2abc(a+b+c)+(ab+bc+ca)^2>6+9=15.$

Ferid.---. wrote: We can find easily $$S=\sum_{cyc} \frac{a^6}{(a-b)(a-c)}= p^2(p^2-3q)+2pr+q^2,$$where $p=a+b+c,q=ab+bc+ca,r=abc,$ and $p^2\ge 3q.$ Then $S=p^2(p^2-3q)+2pr+q^2> 2pr+q^2=2abc(a+b+c)+(ab+bc+ca)^2>6+9=15.$ Very nice. $$\sum_{cyc} \frac{a^{12}}{(a-b)(a-c)}=f(p,q,r)>?$$where $f(p,q,r)=?$ Thank Ferid.---.

anantmudgal09 wrote: Let $a,b,c$ be distinct positive real numbers with $abc=1$. Prove that $$\sum_{\text{cyc}} \frac{a^6}{(a-b)(a-c)}>15.$$ my friend shaxzod's solution We have $$\sum_{\text{cyc}} \frac{a^6}{(a-b)(a-c)}=a^4+b^4+c^4+b^3a+a^3b+a^3c+c^3a+b^3c+c^3b+a^2b^2+a^2c^2+b^2c^2+a+b+c$$from AM-GM we have $LHS >15$.We are done

anantmudgal09 wrote: Let $a,b,c$ be distinct positive real numbers with $abc=1$. Prove that $$\sum_{\text{cyc}} \frac{a^6}{(a-b)(a-c)}>15.$$ Let $a,b,c$ be distinct positive real numbers with $abc=1$. Prove that $$\frac{a^6}{(a-b)(a-c)}+\frac{b^6}{(b-c)(b-a)}+\frac{c^6}{(c-a)(c-b)}> 5(a+b+c)$$

By Cauchy-Schwarz, \[\sum_{cyc} \frac{a^6}{(a-b)(a-c)}\geq \frac{(a^3+b^3+c^3)^2}{\sum_{cyc} (a-b)(a-c)} >15 \iff (a^3+b^3+c^3)^2> 15\left(\sum_{cyc} (a-b)(a-c)\right) \iff 45(ab+bc+ca)+(a^3+b^3+c^3)^2> 15(a+b+c)^2.\]Let $p=a+b+c, q=ab+bc+ca, r=abc=1$, we have to prove \[45q+(p^3-3pq+3)^2>15p^2 \iff 9p^2q^2+(45-18p-6p^4)q+p^6+6p^3-15p^2+9>0\]since this is a quadratic equation in $q$, the discriminant is surprisingly \[\Delta = (45-18p-6p^4)^2-4(9p^2)(p^6+6p^3-15p^2+9)=405(5-4p)\]and our inequality holds iff $\Delta = 405(5-4p)<0 \iff p>\frac{5}{4}$ which is true since $p\geq 3 >\frac{5}{4}$ by AM-GM.$\blacksquare$

Lol quite easy!! anantmudgal09 wrote: Let $a,b,c$ be distinct positive real numbers with $abc=1$. Prove that $$\sum_{\text{cyc}} \frac{a^6}{(a-b)(a-c)}>15.$$ Its easy to see after a bash that the expression is just $$\sum a^4+\sum a^2b^2+\sum ab(a^2+b^2)+abc(\sum a)\geq 3\sqrt[3]{a^4b^4c^4}+3\sqrt[3]{a^4b^4c^4}+6\sqrt[6]{a^8b^8c^8}+3abc\sqrt[3]{abc}=3+3+6+3=15$$with equility iff $a=b=c=1$ but since $a,b,c$ are distinct thus equility isnt attained.$\square$

Keith50 wrote: By Cauchy-Schwarz, \[\sum_{cyc} \frac{a^6}{(a-b)(a-c)}\geq \frac{(a^3+b^3+c^3)^2}{\sum_{cyc} (a-b)(a-c)} >15 \iff (a^3+b^3+c^3)^2> 15\left(\sum_{cyc} (a-b)(a-c)\right) \iff 45(ab+bc+ca)+(a^3+b^3+c^3)^2> 15(a+b+c)^2.\]Let $p=a+b+c, q=ab+bc+ca, r=abc=1$, we have to prove \[45q+(p^3-3pq+3)^2>15p^2 \iff 9p^2q^2+(45-18p-6p^4)q+p^6+6p^3-15p^2+9>0\]since this is a quadratic equation in $q$, the discriminant is surprisingly \[\Delta = (45-18p-6p^4)^2-4(9p^2)(p^6+6p^3-15p^2+9)=405(5-4p)\]and our inequality holds iff $\Delta = 405(5-4p)<0 \iff p>\frac{5}{4}$ which is true since $p\geq 3 >\frac{5}{4}$ by AM-GM.$\blacksquare$ Your solution is wrong, since C-S cannot be applied on negatives.

A solution with Baris Koyuncu, Hakan Gokdogan, Melek Gungor and Sevket Onur Yilmaz. Let’s prove a more general statement, $S_n=\sum_{cyc} \frac{a^n}{(a-b)(a-c)}\geq \binom{n}{2}$. For convention, $S_0=S_1=0= \binom{1}{2}= \binom{0}{2} $ and $S_2=1= \binom{2}{2}$. For $n\geq3$, \begin{align*} \sum_{cyc} a^n(b-c) &= ab(a^{n-1}-b^{b-1})-c(a^n-b^n)+c^n(a-b) \\ &= (a-b)(aba^{n-2}-cba^{n-2}+ab^2a^{n-3}-cb^2a^{n-3}+ \dots ab^{n-1}-cb^{n-1}+c^n-ca^{n-1})\\ &= (a-b)(c-a)(ca^{n-2}-ba^{n-2}+c^2a^{n-3}-b^2a^{n-3}+ \dots c^{n-1}-b^{n-1}) \\ &= (a-b)(c-a)(c-b)\left(a^{n-2}+a^{n-3}(c+b)+a^{n-4}(c^2+cb+b^2)+ \dots a(c^{n-3}+c^{n-2}b+\dots+b^{n-3})\right)\\ &= -(a-b)(b-c)(c-a)(\sum_{\tiny{i+j+k=n-2}}a^ib^jc^k)\\ \end{align*}Then $S_n=\sum_{cyc} \frac{a^n}{(a-b)(a-c)}= \sum_{\tiny{i+j+k=n-2}}a^ib^jc^k $. Since there are $\binom{n}{2}$ nonnegative integer triplets $(i,j,k)$ such that $i+j+k=n-2$, by AM-GM and $abc=1$, $\sum_{\tiny{i+j+k=n-2}}a^ib^jc^k \geq \binom{n}{2}$. The only equality case when $a=b=c$ which is impossible here. $S_n> \binom{n}{2}$.