I think we can do this: $\dfrac{a^8}{a^2c^2+2a^2b^3}$ then Cauchy-Schwarz Maybe it works.

@above

$\frac{a^6}{c^2+2b^3} + \frac{b^6}{a^2+2c^3} + \frac{c^6}{b^2+2a^3} \geq \frac{(a^3+b^3+c^3)^2}{2(a^3+b^3+c^3)+(a^2+b^2+c^2)}$ so we need to show $(a^3+b^3+c^3)^2\geq 2(a^3+b^3+c^3)+(a^2+b^2+c^2)$ $3(a^2+b^2+c^2)\geq (a+b+c)^2=9 \Rightarrow a^2+b^2+c^2\geq 3$ $9(a^3+b^3+c^3)\geq (a+b+c)^3=27 \Rightarrow a^3+b^3+c^3\geq 3$ $(a^3+b^3+c^3)^2=\frac{2(a^3+b^3+c^3)^2}{3}+\frac{(a^3+b^3+c^3)^2}{3}\geq 2(a^3+b^3+c^3)+\frac{(a^2+b^2+c^2)^3}{9}\geq 2(a^3+b^3+c^3)+(a^2+b^2+c^2)$ so we are done.

MarkBcc168 wrote: For all positive real numbers $a,b,c$ with $a+b+c=3$, prove the inequality $$\frac{a^6}{c^2+2b^3} + \frac{b^6}{a^2+2c^3} + \frac{c^6}{b^2+2a^3} \geq 1.$$ $$a^3+b^3+c^3\geq a^2+b^2+c^2\geq a+b+c=3$$$$\frac{a^6}{c^2+2b^3} + \frac{b^6}{a^2+2c^3} + \frac{c^6}{b^2+2a^3} \geq \frac{(a^3+b^3+c^3)^2}{a^2+b^2+c^2+2(a^3+b^3+c^3)} \geq \frac{1}{3}(a^3+b^3+c^3)\geq1.$$https://artofproblemsolving.com/community/c6h1843616p19622329

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