If $(a^5-a^3+3)(b^5-b^3+3)(c^5-c^3+3)\ge 3(a+b+c)^2 \quad$ Prove that $a^5-a^3\ge a^2-1$.

Question

If $$(a^5-a^3+3)(b^5-b^3+3)(c^5-c^3+3)\ge 3(a+b+c)^2 \quad a\ge b\ge c. $$ Prove that $a^5-a^3\ge a^2-1$.

here is my attempt $$a^5-a^3\ge a^2-1$$ is equivalent to $$(a^3-1)(a^1-1)\ge 0$$ or $a\ge -1$ so by their symmety $a,b,c\ge -1$

obviously,$a,b,c=-1or 1$,when it take the equal sign

so how to construct inequalities which satisfies conditions above

enter image description here

Answer 1

Your statement is wrong. Try $(a,b,c)=(-2,-2,1)$.

If $a\geq-1$ then $a^5-a^3\geq a^2-1$ because it's $(a-1)^2(a+1)(a^2+a+1)\geq0$, which is obvious.

If you need to prove that $$\prod_{cyc}(a^5-a^3+3)\geq3(a+b+c)^2$$ for all $\{a,b,c\}\subset[-1,+\infty)$ then $a^5-a^3+3\geq a^2+2$ and similar and see my solution here:

Prove that $(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$

Answer 2

After you deduced the following: $$\prod_{a,b,c}(a^5-a^3+3)\geq\prod_{a,b,c}(a^2+2),$$ the rest is duplicate of this question.

Answer 3

@dezdichado But the question you cite is based on $a,b,c\ge 0$

here $a,b,c\in\mathbb R$

and you may confuse the conditions and conclusions