THQ255

Problem. (Tạ Hồng Quảng) Let $a,b,c>0;a+b+c=2.$ Prove$:$

$$ \dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+14(ab+bc+ca)\geqslant 19$$

Solution. (tthnew) First, notice that $$q=ab+bc+ca\leqslant \dfrac{p^2}{3}=\dfrac{4}{3}.$$

The inequality is equivalent to $${\dfrac {{p}^{3}q-{p}^{2}r-3\,p{q}^{2}+33\,qr-38\,r}{r}}\geqslant {\dfrac { \left( {p}^{2}-q \right)  \left( a-b \right)  \left( b-c \right) \left( a-c \right) }{r}} $$

Or $${p}^{3}q-{p}^{2}r-3\,p{q}^{2}+33\,qr-38\,r\geqslant \left( {p}^{2}-q \right) \left( a-b \right)  \left( b-c \right)  \left(a-c \right)  $$

Or $$\left(33q  -42\right) r+2q \left( 4-3q \right)\geqslant \left( 4-q \right)  \left( a-b \right) \left( b-c \right)  \left(a-c\right)$$

• If $0< q < \dfrac{14}{11}$ then $33q -42 \leqslant 0.$ Therefore:
  $$\text{LHS} \equiv f(r) \geqslant f\Big(\dfrac{q^2}{6}\Big)=\dfrac{1}{2} q\left( 11{q}^{2}-26q+16 \right) \geqslant 0,$$
  
• If $\dfrac{14}{11} \leqslant q \leqslant \dfrac{4}{3}$ clearly the left side is non-negative.
  
  

After squaring both side, need to prove$:$ $$\left( 1116\,{q}^{2}-2988\,q+2196 \right) {r}^{2}+ \left( -432\,{q}^{ 3}+1352\,{q}^{2}-1504\,q+512 \right) r+4\,{q}^{5} \geqslant 0.$$
Note that: $$1116\,{q}^{2}-2988\,q+2196  \geqslant 0.$$
If $$-432\,{q}^{ 3}+1352\,{q}^{2}-1504\,q+512  \geqslant 0 \rightarrow \text{Q. E. D}.$$
Else, $$\Delta_r = -64\, \left( 279\,{q}^{3}-873\,{q}^{2}+864\,q-256 \right)  \left( q-1 \right) ^{2} \left( q-4 \right) ^{2} \leqslant 0.$$
Done.