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xapproachesinfinity

  • one year ago

for \(a,b,c \in \mathbb{R^{+}}\) prove that: \[(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge \left (a+b+c \right)^3\]

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  1. freckles
    • one year ago
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    \[(a^5-a^2+3)(b^3-b^2+3)(c^5-c^2+3) \ge (a+b+c )^3\] just trying to retype it having troubles looking at the other one

  2. xapproachesinfinity
    • one year ago
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    i just corrected it :) should look fine now hahah

  3. xapproachesinfinity
    • one year ago
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    b^5 not b^3

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

  5. freckles
    • one year ago
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    like I'm kinda just looking at the case a=b=c right now

  6. freckles
    • one year ago
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    \[g(x)=x^3 \\ g(3x)=(3x)^3=27x^3 \]

  7. freckles
    • one year ago
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    we should be able to show f^3>=g(3x) if this inequality does hold

  8. xapproachesinfinity
    • one year ago
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    hmm interesting! i was trying AM-GM inequality i was interested in this after i saw some questions related to that inequality

  9. freckles
    • one year ago
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    well I don't know if my way is good or not and so far I haven't gotten anywhere quick yet

  10. xapproachesinfinity
    • one year ago
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    i didn't really go that far with am-gm either but i think should work

  11. xapproachesinfinity
    • one year ago
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    anyways gotta go have a paper to do lol

  12. freckles
    • one year ago
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    I have to think on this more later too time to eat

  13. anonymous
    • one year ago
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    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
    • one year ago
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