Prove that: $(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$

Question

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

Answer 1

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:

1. $w^3\rightarrow0^+$;

2. Two variables are equal.

Answer 2

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 $0<a<\sqrt3$.

Answer 3

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