# Math Help - Can you solve this !!

1. ## Can you solve this !!

Prove that
( x^100) / e^(x^70) + e^(x^70) /(x^100) > 2

for every positive x.

Prove that
cos 20 =/= a/b where a and b are integers .

Nice questions i did them i just wanted to see other solutions.

This should be easy for a lot of you i looked around the forum there are a lot of beautiful mathematicians around.

2. Originally Posted by ╔(σ_σ)╝
Prove that
( x^100) / e^(x^70) + e^(x^70) /(x^100) > 2

for every positive x.

Prove that
cos 20 =/= a/b where a and b are integers .

Nice questions i did them i just wanted to see other solutions.

This should be easy for a lot of you i looked around the forum there are a lot of beautiful mathematicians around.
For the first one, let $u = \frac{x^{100}}{e^{x^{70}}}$, then

$\frac{x^{100}}{e^{x^{70}}} + \frac{e^{x^{70}} }{x^{100}} = u + \frac{1}{u}$

so you want to show $u + \frac{1}{u} > 2$ which is to show that

$u^2 - 2u + 1 > 0$ or $(u-1)^2 > 0$ which is true provide that $u \ne 1$. Now u is bounded above (using some calculus) so $u \le u_{max} < 1$ establishing the inequality.

3. Beautiful.

That's what you need to do.

You have a trained eye, or u must have seen this before.

4. Originally Posted by ╔(σ_σ)╝
Beautiful.

That's what you need to do.

You have a trained eye, or u must have seen this before.
I've been doing math for a while

5. Originally Posted by danny arrigo
I've been doing math for a while
Cool

Now about the second one .

Experience won't help.

6. Amazing no one can solve the second one !

7. Originally Posted by ╔(σ_σ)╝
Prove that
cos 20 =/= a/b where a and b are integers .
This is an idea, I did not even try doing this.
By the triple angle identity we have a cubic polynomial equaling $\cos 60 = \frac{\sqrt{3}}{2}$.
Now if we square both sides of the polynomial to make everything rational we just need to argue the polynomial has no rational solutions.

8. Originally Posted by ThePerfectHacker
This is an idea, I did not even try doing this.
By the triple angle identity we have a cubic polynomial equaling $\cos 60 = \frac{\sqrt{3}}{2}$.
Now if we square both sides of the polynomial to make everything rational we just need to argue the polynomial has no rational solutions.
$\cos 60 =/= \frac{\sqrt{3}}{2}$

$\cos 60 = \frac{\sqrt{1}}{2}$

Finish it.

9. Let $x=\cos 20$ then $4x^3 - 3x = \tfrac{\sqrt{3}}{2}$.
If $x$ was rational then $4x^3 - 3x$ would be rational but it is not.
Thus, $x$ must be irrational.

10. Originally Posted by ThePerfectHacker
Let $x=\cos 20$ then $4x^3 - 3x = \tfrac{\sqrt{3}}{2}$.
If $x$ was rational then $4x^3 - 3x$ would be rational but it is not.
Thus, $x$ must be irrational.
Why do you keep saying this ?
$\cos 60 = \frac{\sqrt{3}}{2}$ -- This is wrong

cos 60 = 1/2

11. Originally Posted by ╔(σ_σ)╝
Why do you keep saying this ?
$\cos 60 = \frac{\sqrt{3}}{2}$ -- This is wrong

cos 60 = 1/2
Sorry! That was a trivial error.
But the idea still works because: $4x^3 - 3x = \tfrac{1}{2} \implies 8x^3 - 6x -1 = 0$.
Apply rational roots theorem and conclude there are no rational solutions.