Hint. Given three length-sides $a, b, c$ of a triangle. For $\Delta= \sqrt{s(s- a)(s- b)(s- c)}$, prove : $$abc+ (a- b)(b- c)(c- a)\geqq \frac{8}{3}\sqrt{\sqrt{3}\Delta^{3}} \tag{1}$$ $$abc- (a- b)(b- c)(c- a)\geqq \frac{8}{3}\sqrt{\sqrt{3}\Delta^{3}} \tag{2}$$
Remark. Given four positive numbers $b, c, d, e$ so that $ab+ c= -ad+ e$ (as inequalties above) : $$\therefore\,ab+ c= -ad+ e= \frac{eb+ cd}{b+ d}\geqq 0$$ After using $\lceil$ Ravi-substitution $\rfloor$ and more, we have a problem due to Liu Qian Bao as follow :
Problem. Given three positive numbers $x, y, z$, prove that : $$(xyz+ x^{2}y+ y^{2}z+ z^{2}x)^{4}\geqq \frac{256}{27}(x+ y+ z)^{3}x^{3}y^{3}z^{3}$$
I think the best way here is to prove $(1)$ and $(2)$ by $uvw$ because they're $cyc$, but hard to square them
Yes, it's true by the $uvw$'s technique.
Indeed, let $x+y+z=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u=t^{\frac{4}{3}}w$.
Thus, we need to prove that $$w^3+\sum_{cyc}x^2y\geq4\sqrt[4]{u^3w^9}$$ or $$2w^3+\sum_{cyc}(x^2y+x^2z)-8\sqrt[4]{u^3w^9}\geq\sum_{cyc}(x^2z-x^2y)$$ or $$9uv^2-w^3-8\sqrt[4]{u^3w^9}\geq(x-y)(y-z)(z-x).$$ Now, since by Maclaurin $$9uv^2-w^3-8\sqrt[4]{u^3w^9}\geq0,$$ it's enough to prove that $$\left(9uv^2-w^3-8\sqrt[4]{u^3w^9}\right)^2\geq\prod_{cyc}(x-y)^2(x-z)^2(y-z)^2$$ or $$\left(9uv^2-w^3-8\sqrt[4]{u^3w^9}\right)^2\geq27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $$27v^6-(45uv^2w^3+36uv^2\sqrt[4]{u^3w^9})v^2+7w^6+27u^3w^3+4w^3\sqrt[4]{u^3w^9}+16\sqrt{u^3w^9}\geq0.$$ Now, by AM-GM $$27v^6+7w^6+27u^3w^3+4w^3\sqrt[4]{u^3w^9}+16\sqrt{u^3w^9}=$$ $$=27v^6+2\cdot\frac{7w^6+27u^3w^3+4w^3\sqrt[4]{u^3w^9}+16\sqrt{u^3w^9}}{2}\geq$$ $$\geq3\sqrt[3]{27v^6\cdot\left(\frac{7w^6+27u^3w^3+4w^3\sqrt[4]{u^3w^9}+16\sqrt{u^3w^9}}{2}\right)^2}.$$ Id est, it's enough to prove that $$3\sqrt[3]{27v^6\cdot\left(\frac{7w^6+27u^3w^3+4w^3\sqrt[4]{u^3w^9}+16\sqrt{u^3w^9}}{2}\right)^2}\geq(45uv^2w^3+36uv^2\sqrt[4]{u^3w^9})v^2$$ or $$(27t^4+16t^2+14t+7)^2\geq4t^4(4t+5)^3$$ or $f(t)\geq0,$ where $$f(t)=2\ln(27t^4+16t^2+4t+7)-4\ln{t}-3\ln(4t+5)-2\ln2.$$ Now, $$f'(t)=\frac{2(108t^3+32t+4)}{27t^4+16t^2+4t+7}-\frac{4}{t}-\frac{12}{4t+5}=$$ $$=\frac{4(t-1)(27t^4+162t^3+114t^2+94t+35)}{t(4t+5)(27t^4+16t^2+4t+7)},$$ which gives $t_{\min}=1$ and since $f(1)=0$, we are done!