xapproachesinfinity
  • xapproachesinfinity
for \(a,b,c \in \mathbb{R^{+}}\) prove that: \[(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge \left (a+b+c \right)^3\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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freckles
  • 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
xapproachesinfinity
  • xapproachesinfinity
i just corrected it :) should look fine now hahah
xapproachesinfinity
  • xapproachesinfinity
b^5 not b^3

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freckles
  • 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 ...
freckles
  • freckles
like I'm kinda just looking at the case a=b=c right now
freckles
  • freckles
\[g(x)=x^3 \\ g(3x)=(3x)^3=27x^3 \]
freckles
  • freckles
we should be able to show f^3>=g(3x) if this inequality does hold
xapproachesinfinity
  • xapproachesinfinity
hmm interesting! i was trying AM-GM inequality i was interested in this after i saw some questions related to that inequality
freckles
  • freckles
well I don't know if my way is good or not and so far I haven't gotten anywhere quick yet
xapproachesinfinity
  • xapproachesinfinity
i didn't really go that far with am-gm either but i think should work
xapproachesinfinity
  • xapproachesinfinity
anyways gotta go have a paper to do lol
freckles
  • freckles
I have to think on this more later too time to eat
anonymous
  • 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.
anonymous
  • anonymous
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