THQ256

Problem. (Tạ Hồng Quảng) For $a,b,c>0;a+b+c=1.$ Prove$:$ 

$$ \dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+28(ab+bc+ca)^2\geqslant  3 $$

Solution.
We write the inequality as $$( 56{q}^{2}+5q-7 ) r + q(1-3q) \geqslant \left( 1-q \right)  \left( a-b \right)  \left( b-c \right)(a-c)$$

• If $\displaystyle{\dfrac {3\sqrt{177}}{112}}-{\dfrac {5}{112}} \leqslant q \leqslant \dfrac{1}{3}$ then clearly the left side is non-negative.

• If $0 < q < {\dfrac {3\sqrt{177}}{112}}-{\dfrac {5}{112}}$ then $ 56{q}^{2}+5q-7 <0.$
  We have$:$ $$\text{LHS} \equiv f(r) \geqslant f\Big(\dfrac{q^2}{3}\Big)=\dfrac{1}{3} q \left( 56\,{q}^{3}+5{q}^{2}-16q+3 \right)\geqslant 0.$$.
  

So the left side is non-negative. After squaring$,$ need to prove$:$ $$\left( 3136\,{q}^{4}+560\,{q}^{3}-732\,{q}^{2}-124\,q+76 \right) {r}^ {2}+ \left( 64\,{q}^{3}+92\,{q}^{2}-40\,q-336\,{q}^{4}+4 \right) r+4\, {q}^{5}\geqslant 0.$$

• If $64\,{q}^{3}+92\,{q}^{2}-40\,q-336\,{q}^{4}+4 \geqslant 0 \rightarrow \text{Q. E. D}.$

• Else we have $$\Delta_r = -16\, \left( 7\,q-1 \right)  \left( 28\,{q}^{4}+16\,{q}^{3}-8\,{q}^{2} -3\,q+1 \right)  \left( 4\,q-1 \right) ^{2} \left( q-1 \right) ^{2} \leqslant 0.$$

Done.