Maximization: find max value of (x^4-x^2)/(x^6+2x^3-1) for x > 1

[Otis]

I've tried working this exercise a few ways, but I haven't yet been able to obtain the answer without using a calculator. (My sabbatical isn't helping any, I'm sure...)

 

[daniellionyang]

I've tried working this exercise a few ways, but I haven't yet been able to obtain the answer without using a calculator. (My sabbatical isn't helping any, I'm sure...)

Ok... Here I go!
(x^4-x^2)/(x^6+2x^3-1)
=>(x^3(x-1/x))/(x^3(x^3+2-1/x^3))
Cross out x^3
=>(x-1/x)/(x^3-1/x^3+2)
let x-1/x=a.
Then, a^3=x^3-3x+3/x-1/x^3
and a^3+3a=x^3-1/x^3
(we can do this since x>1)
=>a/(a^3+3a+2)
Weighted AM-GM on a^3,a,a,a,1,1
=> a^3+3a+2<=(less than or equal to)6*6throot(a^3*a*a*a*1*1)=6a.
=>a/6a=1/6. QED.

I guess you can use calculus to avoid AM-GM, but it gets ugly
http://www.artofproblemsolving.com/wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality

 

[Otis]

Weighted AM-GM on a^3,a,a,a,1,1
--> a^3+3a+2<=6*6throot(a^3*a*a*a*1*1)=6a.
-->a/6a=1/6. QED.

Nice, but way over my head. :cool:

(I've never heard of an Inequality course, either.)