Helpful lemma for the homogeneous inequality

Let $p=a+b+c=1,q=ab+bc+ca,r=abc.$ Since $$ {q}^{2} \left( {p}^ {2}-4\,q \right) -2\,p \left( 2\,{p}^{2}-9\,q \right) r-27\,{r}^{2}=(a-b)^2(b-c)^2(c-a)^2 \geqslant 0.$$ Or $${q}^{2} \left( 1-4\,q \right)-2\, \left( 2-9\,q \right) r -27\,{r}^{2}\geqslant 0$$ Therefore $$-{\frac {2}{27}}+\frac13\,q-{\frac {2}{27}}\,\sqrt { \left( 1-3\,q \right) ^{3}} \leqslant r \leqslant -{\frac {2}{27}}+\frac13\,q+{\frac {2}{27}}\,\sqrt { \left( 1-3\,q \right) ^{3}}.$$ Let $t=\sqrt{1-3q} \quad \Big(t\geqslant 0 \Big) \Rightarrow q=\frac{1-t^2}{3}.$ We have$:$ $$\boxed{\dfrac{1}{27} \left( 1-2t \right) \left( 1+t \right) ^{2} \leqslant r\leqslant \dfrac{1}{27} \left( 2\,t+1 \right) \left( t-1\right) ^{2}}$$ Note. If $a,b,c\geqslant 0$ then we also have $t \in [\,0,1\,].$ Now we see that it's same as Bổ Đề Chặn Tích by Vo Quoc Ba Can :)