<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] Thanks for helping me, I also have a solution after many days of thinking
$$[9(ab+bc+ca)]^2=27(a^2+b^2+c^2)(ab+bc+ca)^2$$
By AM-GM, we have:
$$27(a^2+b^2+c^2)(ab+bc+ca)(ab+bc+ca)\le(a+b+c)^6$$
$$\to 9(ab+bc+ca)\le(a+b+c)^3 $$
The problem is:
$$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge(a+b+c)^3$$
By AM-GM, we have:
$$a^5-2a+4=\frac{1}{5}(a^5+a^5+1+1+1)-2a+1+\frac{1}{5}(a^5+a^5+a^5+1+1+10)\geq (a-1)^2+\frac{1}{5}(5a^3+10)\geq a^3+2$$
$$\to (a^5-2a+4)(b^5-2b+4)(c^5-2c+4) \ge (a^3+2)(b^3+2)(c^3+2) $$
The problem is:
$$(a^3+2)(b^3+2)(c^3+2) \ge (a+b+c)^3 $$
Which is true by Holder
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