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

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freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[(a^5a^2+3)(b^3b^2+3)(c^5c^2+3) \ge (a+b+c )^3\] just trying to retype it having troubles looking at the other one

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i just corrected it :) should look fine now hahah

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0b^5 not b^3

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x)=x^5x^2+3 \\ f(x) >0 \text{ for } x>0 \\ \text{ and we have} \\ g(x)=x^3 \\ g(x) > 0 \text{ for } x>0 \\ \] \[(f(x))^3=(x^5x^2+3)^3 \\ =(x^2(x^31)+3)^3 \\ =(x^2(x^31))^3+3(x^2(x^31))^2(3)+3(x^2(x^31))(3)^2+3^3 \\ =x^5(x^31)^3+9x^4(x^31)^2+27x^2(x^31)+27 \] I'm just thinking here ...

freckles
 one year ago
Best ResponseYou've already chosen the best response.0like I'm kinda just looking at the case a=b=c right now

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[g(x)=x^3 \\ g(3x)=(3x)^3=27x^3 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0we should be able to show f^3>=g(3x) if this inequality does hold

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hmm interesting! i was trying AMGM inequality i was interested in this after i saw some questions related to that inequality

freckles
 one year ago
Best ResponseYou've already chosen the best response.0well I don't know if my way is good or not and so far I haven't gotten anywhere quick yet

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i didn't really go that far with amgm either but i think should work

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0anyways gotta go have a paper to do lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I have to think on this more later too time to eat

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not 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^5w^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.
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