Prove the inequality

[Math Tutor]

Prove the inequality:
[tex]6^{\frac{1}{3}}+6^{-\frac{1}{3}}>2[/tex]

[Guest]

x + 1/x >= 2, (A.M.-G.M.) => Q.E.D.

Y0UrShAD0W (YS)

[Guest]

For this case a_1 <> a_2, a_1, a_2 >0 then x + 1/x >2 (according to A.M.G.M.)

YS

[Guest]

Or f`(c_1) < f`(c_2) - study in 6^x function in [-t,0] and [0,t] => Q.E.D.

y0urshad0w

[Guest]

Multiply by 6^(1/3) in both sides => (6^(1/3)-1)^2>0


YS

[Math Tutor]

How do you calculated without a calculator or any online tool that (6^(1/3)-1)^2>0

About x + 1/x >= 2 it is true but we have a strict inequality.

[Gus123456789]

Let's see, I think the key to this is the fact that we can rewrite this as a more general inequality.
[tex]6^{\frac{1}{3}}+6^{\frac{-1}{3}}>2[/tex]

[tex]6^{x}+6^{-x}>2[/tex] (provided variable x such that x is not zero)
Notice that the let side has the form of a function f(x). I think there is an absolute minimum at the limit near x=0. We can prove that with minimal calculus. There can only be a minimum or maximum of a function where the first derivative is 0 or undefined. Our function f(x) is undefined or zero only at x=0, so the only possible minimum is there. Because it is undefined, so is its derivative. We need 3 points to confirm a minimum. We need the limit of f(x) at the point of interest, and 2 values of f(x) close to the point of interest on opposite sides. If both side values are greater than our point of interest value, then it is a minimum. Our point of interest limit value is
[tex]6^{0}+6^{0}=2[/tex]
We pick x=1 and x=-1 for the side values.
[tex]6^{1}+6^{-1}=6^{(-1)}+6^{-(-1)}=6+\frac{1}{6}[/tex] which confirms that the limit near x=0 is a minimum.
Because this minimum is not actually a pat of the function, just a limit, we can say for certain that for every possible x, f(x)>2. And that includes our original inequality. Therefore, [tex]6^{\frac{1}{3}}+6^{\frac{-1}{3}}>2[/tex]

[Guest]

Study the monotony of

f(x) = a^x + a^(b/x) in (0, oo), a>0, b>0, a<>1


YS.

[Guest]

Mrs. Teacher what exactly should i prove here ?

"How do you calculated without a calculator or any online tool that (6^(1/3)-1)^2 > 0

YS

[Guest]

how about that
[tex]6^{\frac{1}{3}}=y[/tex]

[tex]y^2-2y+1>0[/tex]

[tex]y>0[/tex]

[tex](y-1)^2>0[/tex] which applies for every y but 1
and since [tex]6^{\frac{1}{3}}\ne 1[/tex] ... ;]

[Guest]

pretty much the same as YS's solution ;]

[Math Tutor]

I think that the best is the last solution.

[Guest]

Mrs. Teacher i think you try to make a difference between 2 identical things and say that
one is better than the other. Interesting approach but less logical, though.

YS

[Math Tutor]

You are right but his explanation is better.

[Guest]

Mrs. Teacher, let me remind you that you referred to the solution, and not the level explanation of the solutions.
In terms of solutions the 2 answers are identical as approach. I`ll never have as aim to reach a high explanations
level since i have no interes to do that.

YS.