<s>[INST] Prove that: $(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$ Let $a,b,c>0$ satisfy $a^2+b^2+c^2=3$ . Prove that: $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$$
My idea is to use a well-known inequality (We can prove by Schur) $$(a^2+2)(b^2+2)(c^2+2)\ge 9(ab+bc+ca)$$ and the problem is prove  $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge(a^2+2)(b^2+2)(c^2+2)$$
Anyone have a better idea ? please help me
 [/INST] Another way: use uvw https://math.stackexchange.com/tags/uvw/info
Since $a^5-2a+4\geq4a^2-5a+4,$ it's enough to prove that:
$$\prod_{cyc}(4a^2-5a+4)\geq9(ab+ac+bc),$$ which is $f(w^3)\geq0,$ where $f$ increases. I got here:
$$f'(w^3)=128w^3+684u+115-240v^2>0.$$
Thus, it's enough to prove the last inequality in two cases:

*

*$w^3\rightarrow0^+$;


*Two variables are equal.
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