<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] Since $ab+ac+bc\leq a^2+b^2+c^2=3,$ it's enough to prove that:
$$\prod_{cyc}(a^5-2a+4)\geq27$$ or
$$\sum_{cyc}\left(\ln(a^5-2a+4)-\ln3\right)\geq0$$ or
$$\sum_{cyc}\left(\ln(a^5-2a+4)-\ln3-\frac{1}{2}(a^2-1)\right)\geq0.$$
Now, prove that $$\ln(a^5-2a+4)-\ln3-\frac{1}{2}(a^2-1)\geq0$$ for any $0<a<\sqrt3$.
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