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

score 0 · Answer 1 · answered Jul 26 '17 at 02:49

0

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.

answered Jul 26 '17 at 02:49

dezdichado

13,888

score 0 · Answer 2 · answered Jul 26 '17 at 03:15

0

@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

answered Jul 26 '17 at 03:15

lurk lurk

9

that's because it suffices to assume that. use $a^2 = (-a)^2.$ and $a+b+c\leq |a|+|b|+|c|.$ – dezdichado Jul 26 '17 at 04:08
by using $\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right) \ge 3\left(a^2+b^2+1\right)$ ,how to prove $a^5-a^3\ge a^2-1$ @dezdichado – lurk lurk Jul 26 '17 at 04:22

Michael Rozenberg · Accepted Answer · 2017-07-26T05:01:15.023

0

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

edited Jul 26 '17 at 05:01

answered Jul 26 '17 at 04:27

Michael Rozenberg

194,933

if$ a,b,c\ge 0$,then how to do it – lurk Jul 26 '17 at 04:45
@lurk lurk vice account I fixed my post. See now please. – Michael Rozenberg Jul 26 '17 at 04:55
so how to prove that $a^5-a^3\ge a^2-1$ if $a,b,c\ge -1$ – lurk Jul 26 '17 at 05:00
@lurk lurk vice account See the second line of my solution. I fixed something. – Michael Rozenberg Jul 26 '17 at 05:02
if $a\ge b\ge c$ then how – lurk Jul 26 '17 at 05:10
@lurk lurk vice account Can you say a full statement? – Michael Rozenberg Jul 26 '17 at 05:11
I am trying to find the missing condition – lurk Jul 26 '17 at 05:14
@lurk lurk vice account Where did you take this problem? – Michael Rozenberg Jul 26 '17 at 05:17
https://i.stack.imgur.com/9Rce2.png – lurk Jul 26 '17 at 05:21
@lurk lurk vice account Can you write in Chines without pictures? I'll can use the google-translate. – Michael Rozenberg Jul 26 '17 at 05:37
the second line in the picture is " 证明（xxx数学工作室） " ，words in brackets is the source of the question .As you said above ,it is wrong ,so I guess it lacks some conditions (like $a\ge b\ge c$ ) – lurk Jul 26 '17 at 05:45

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 $a^5-a^3\ge a^2-1$.

3 Answers3