Let me give you my idea First we can prove $$\dfrac{ab+bc+ca}{(a+b)(b+c)(c+a)}<\dfrac{1}{a+b-2c}.$$which is quite obvious for positive real a, b, c We can suppose a>b>c and x=a-b, y=b-c Then we only need to prove $$\dfrac{1}{x+2y}<\dfrac17(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x+y}).$$We then suppose t=y/x We only need to prove3<2t+1/t+t/(1+t) Then we get 2t^3+1>2t using AM-GM, 2t^3+1/2+1/2>=3t(1/2)^(1/3)>2t The proof is complete!

My proof is use $a+b=z , b+c=x , c+a=y$ and WLOG set $z\ge y\ge x$ with $z-y=m , y-x=n$ where$x , y , z , m , n$ are all positive. Then I expanded it all out; my sol is definitely not as good as shiraikuroko's.

Henry_2001 wrote: Let $a,b,c$ be pairwise distinct positive real, Prove that$$\dfrac{ab+bc+ca}{(a+b)(b+c)(c+a)}<\dfrac17(\dfrac{1}{|a-b|}+\dfrac{1}{|b-c|}+\dfrac{1}{|c-a|}).$$ Let $a,b,c $ be distinct nonnegative real numbers such that $a+b+c=3$. Prove that$$\frac{1}{|a-b|}+\frac{1}{|b-c|}+\frac{1}{|c-a|}>\frac{12}{5} $$Let $a,b,c$ be positive numbers such that $a+b+c=1,(a-b)(b-c)(c-a)\neq0$,prove that $$\frac{1}{|a-b|}+\frac{1}{|b-c|}+\frac{1}{|c-a|}>{\frac{36}{5}}$$

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