<s>[INST] Proving that $a^{4}+b^{3}+c^{2}\ge a^{3}+b^{3}+c$ for positive real $a,b,c$, $abc\ge 1$ I have been able to simplify the problem. Since $a^{4}\ge a^{3}$ iff $a\ge 1$ (and analogously for $b,c$), $a^{4}+b^{3}+c^{2}\ge a^{3}+b^{3}+c$ holds iff $a+b+c\ge 3$. It is thus sufficient to prove $a+b+c\ge 3$.
I am not sure how to proceeds from there, however.
All insights are really appreciated.
 [/INST] I think maybe the question should be $a^4+b^3+c^2\ge a^3+b^2+c$.
By AM-GM,
$a^4+a^4+a^4+1\ge 4a^3\Rightarrow a^4\ge \frac{4a^3-1}{3}$
Similarly, $b^3\ge \frac{3b^2-1}{2}$ , $c^2\ge 2c-1$
Hence, $a^4+b^3+c^2\ge \frac{1}{6}(8a^3+9b^2+12c-11)$
So we only need to prove that
$8a^3+9b^2+12c-11\ge 6a^3+6b^2+6c$ , which is equivalent to
$2a^3+3b^2+6c\ge 11$
By AM-GM, $2a^3+3b^2+6c=a^3+a^3+b^2+b^2+b^2+c+c+c+c+c+c\ge 11\sqrt[11]{a^6b^6c^6}\ge 11$
Hence, we are done.
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