Prove $\sum\limits_{cyc}\frac{(b+c-a)^4}{a(a+b-c)}\geq ab+bc+ca$

Question

if $a,b,c$ are positive real numbers,Prove:$$\sum\limits_{cyc}\frac{(b+c-a)^4}{a(a+b-c)}\geq ab+bc+ca$$ Additional info: Problem should be solved with AM-GM inequality only.

Things i have tried so far: Using AM-GM I can say $$\frac{((b+c-a)^2)^2}{a(a+b-c)}+a(a+b-c)\geq 2 (b+c-a)^2$$

is true.So i can conclude that:$$\sum\limits_{cyc}\frac{(b+c-a)^4}{a(a+b-c)}+\sum\limits_{cyc}a(a+b-c)\geq \sum \limits_{cyc}2(b+c-a)^2$$

So if I prove that $$\sum \limits_{cyc}2(b+c-a)^2-\sum\limits_{cyc}a(a+b-c)\geq ab+ac+bc$$ then the problem is solved.I stuck in here.I tried to write left side in expanded form but i was unsuccessful at the end to prove it.

UPDATE

I expanded left side and i got this:$$\sum \limits_{cyc}2(b+c-a)^2-\sum\limits_{cyc}a(a+b-c) = 5a^2+5b^2+5c^2-4ab-4bc-4ca$$

so I can rewrite the last inequality as $a^2+b^2+c^2\geq ab+bc+ca$ which is obviously true

Answer 1

Check out again the last step, it become $a^2+b^2+c^2\geq ab+bc+ca$, and now that is easy

Answer 2

Your inequality is wrong! Try $a+b-c\rightarrow0^-$.