xapproachesinfinity one year ago for $$a,b,c \in \mathbb{R^{+}}$$ prove that: $(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge \left (a+b+c \right)^3$

1. freckles

$(a^5-a^2+3)(b^3-b^2+3)(c^5-c^2+3) \ge (a+b+c )^3$ just trying to retype it having troubles looking at the other one

2. xapproachesinfinity

i just corrected it :) should look fine now hahah

3. xapproachesinfinity

b^5 not b^3

4. freckles

$f(x)=x^5-x^2+3 \\ f(x) >0 \text{ for } x>0 \\ \text{ and we have} \\ g(x)=x^3 \\ g(x) > 0 \text{ for } x>0 \\$ $(f(x))^3=(x^5-x^2+3)^3 \\ =(x^2(x^3-1)+3)^3 \\ =(x^2(x^3-1))^3+3(x^2(x^3-1))^2(3)+3(x^2(x^3-1))(3)^2+3^3 \\ =x^5(x^3-1)^3+9x^4(x^3-1)^2+27x^2(x^3-1)+27$ I'm just thinking here ...

5. freckles

like I'm kinda just looking at the case a=b=c right now

6. freckles

$g(x)=x^3 \\ g(3x)=(3x)^3=27x^3$

7. freckles

we should be able to show f^3>=g(3x) if this inequality does hold

8. xapproachesinfinity

hmm interesting! i was trying AM-GM inequality i was interested in this after i saw some questions related to that inequality

9. freckles

well I don't know if my way is good or not and so far I haven't gotten anywhere quick yet

10. xapproachesinfinity

i didn't really go that far with am-gm either but i think should work

11. xapproachesinfinity

anyways gotta go have a paper to do lol

12. freckles

I have to think on this more later too time to eat

13. anonymous

Not sure if this is correct at all, but I graphed a representation of each side of the inequality to compare them, as can be seen in the attached screenshot. The expression on the left is the product of the same "function" evaluated three times at a, b, then c. So, I defined a function f(w) = w^5-w^2+3, chose values for constants a, b, and c, and then graphed the product f(x+a)f(x+b)f(x+c). I then graphed y=(x+a+b+c)^3. Clearly, the first equation (in blue in the screenshot) is greater than the second, if the graph is correct, that is. I tried picking different values for a, b, and c, and the first equation was still greater than the second. Again, I don't know if this approach is valid, but it is all I can think of at the moment.

14. anonymous

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