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If $AB=c, BC=a, CA=b$ then we need to show that $$(b-c)^2(a^2+4bc)^2\le 2a^6$$or $$b^6+c^6+3(b^4c^2+b^2c^4)+20b^3c^3\ge 6(b^5c+bc^5).$$WLOG $c=1$. Consider the function $$f(b)=b^6+1+3(b^2+b^4)+20b^3-6(b^5+b).$$Note that $$f'(b)=6b^5+6b+12b^3+60b^2-30b^4-6=6(b+1)(b^2-4b+1)(b^2-2b-1)=$$$$=6(b+1)(b-(2+\sqrt{3}))(b-(2-\sqrt{3}))(b-(1+\sqrt{2}))(b-(1+\sqrt{2})).$$Thus we need to check the inequality $f(b)\ge 0$ only for $b=2+\sqrt{3}, b=2-\sqrt{3}, b=1+\sqrt{2}$ which is not hard to do.

Let $x=\tan\angle{ACB}$. We desire \[ (x-1)^2(x^2+4x+1)^2\le2(x^2+1)^3. \]But observe that \[ (x-1)^2(x^2+4x+1)^2+(x+1)^2(x^2-4x+1)^2=2(x^2+1)^3. \]Indeed, both sides are the unique even polynomial of degree $6$ with leading coefficient $2$ sending $0\mapsto2$, $1\mapsto16$, $i\mapsto0$. The inequality follows from this identity, also implying equality precisely when $\triangle{ABC}$ has acute angles $15^\circ,75^\circ$.