Prove $(\frac{a}{b-c})^2+(\frac{b}{c-a})^2+(\frac{c}{a-b})^2 \geq 2$

Question

If $a, b, c$ are distinct real numbers, prove that

$(\frac{a}{b-c})^2+(\frac{b}{c-a})^2+(\frac{c}{a-b})^2 \geq 2$

I thought of using AM-GM but that is surely not getting me anywhere ( Maybe some step before i can do AM-GM ?)

I thought of applying AM-GM but that clearly doesnt give anything. Then i thought if a assuming \txtWloga>b>c, that would make the first time greater than 1 so if only i could show that the sumof the other two terms is ≥1 i'd be done but ofc i couldn't progress further from there.

Thanks.

Answer 1

$$\sum_{cyc}\frac{a^2}{(b-c)^2}=\left(\sum_{cyc}\frac{a}{b-c}\right)^2-2\sum_{cyc}\frac{ab}{(b-c)(c-a)}=$$ $$=\left(\sum_{cyc}\frac{a}{b-c}\right)^2-2\sum_{cyc}\frac{ab(a-b)}{\prod\limits_{cyc}(a-b)}=\left(\sum_{cyc}\frac{a}{b-c}\right)^2+2\geq2.$$

Answer 2

Note $$ \sum a(2a-b-c)=2(a^2+b^2+c^2-ab-bc-ca),$$ and $$\sum (b-c)^2(2a-b-c)^2=2(a^2+b^2+c^2-ab-bc-ca)^2.$$ Therefore, according to the Cauchy-Schwarz inequality we have $$\sum \frac{a^2}{(b-c)^2} \geqslant \frac{\left[a(2a-b-c)+b(2b-c-a)+c(2c-a-b)\right] ^2}{(b-c)^2(2a-b-c)^2+(c-a)^2(2b-c-a)^2+(a-b)^2(2c-a-b)^2}=2.$$