help prove inequality when $3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right)$

Question

Given the positive real numbers $a,b,c$ satisfy $3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right)$. Prove that $$\frac{{{a^2}}}{{b + 2c}} + \frac{{{b^2}}}{{c + 2a}} + \frac{{{c^2}}}{{a + 2b}} \ge \frac{{16}}{{27}}$$

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AM-GM we have $\frac{{{a^2}}}{{b + 2c}} +\dfrac{a^2(b+2c)}{9} \geq \dfrac{2}{3}a^2$ $=>P \geq \dfrac{2}{3}(a^2+b^2+c^2)-\dfrac{1}{9}(a^2b+b^2c+c^2a+2(ca^2+ab^2+bc^2))$ Cauchy-Schwarz $a^2b+b^2c+c^2a \leq \sqrt{(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)} \leq \dfrac{1}{\sqrt{3}}(a^2+b^2+c^2)\sqrt{a^2+b^2+c^2}$ And $ab^2+bc^2+ca^2 \leq \dfrac{1}{\sqrt{3}}(a^2+b^2+c^2)\sqrt{a^2+b^2+c^2}$ $=>P \geq \dfrac{2}{3}t^2-\dfrac{\sqrt3}{9}t^3$ $3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right)$=>$3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right) \geq (a^2+b^2+c^2)^2+10$ $=>2 \leq t^2 \leq 5$ We have $f(t)=\dfrac{2}{3}t^2-\dfrac{\sqrt3}{9}t^3$

$f'(t)=\dfrac{4t}{3}-\dfrac{\sqrt3t^2}{3}$ We have $Minimize=\dfrac{\sqrt{6}}{3}$ but "=" ... (But I can't continue, help me)

Answer 1

By C-S $$\sum_{cyc}\frac{a^2}{b+2c}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(b+2c)}=\frac{a+b+c}{3}.$$ Thus, it remains to prove that $$a+b+c\geq\frac{16}{9}$$ or $$\sum_{cyc}\left(a-\frac{16}{27}+\frac{1}{2}\left(3a^4-7a^2+\frac{10}{3}\right)\right)\geq0$$ or $$\sum_{cyc}(81a^4-189a^2+54a+58)\geq0,$$ which is true because $$81a^4-189a^2+54a+58=27(3a^4-7a^2+2a+2)+4=27(a-1)^2(3a^2+6a+2)+4>0.$$ Done!