Prove that, for all x1,

2√x 3 - 1/x

Question

Prove that, for all x>1,

2√x > 3 - 1/x

anonymous · Answer

Hi

agentjamesbond007 · Answer

Hello

anonymous · Answer

I think i might know how to solve this problem

anonymous · Answer

First we get all x's on one side

agentjamesbond007 · Answer

I tried to isolate x to one side. I can't seem to find a way.

anonymous · Answer

now is the x under the 1 or is the x under the 3-1?

agentjamesbond007 · Answer

the x is under the 1

anonymous · Answer

ok give me a moment to work it out on paper

anonymous · Answer

I think your statement about x>1 is false. 
For example,$$\frac{ 1 }{ 2 }$$ is not greater than 1 but the equation is still true.

agentjamesbond007 · Answer

So would that be a way to prove it?

anonymous · Answer

Prove it false yes.

anonymous · Answer

no you will need to get x by itself

anonymous · Answer

to actually prove it

agentjamesbond007 · Answer

oh , ok.

anonymous · Answer

I don't know if it the sign was suppose to be written as ≥.

If you plug in 1, you will get 2 > 2, which is false.

agentjamesbond007 · Answer

I keep getting into a loop when I divide 2 to both sides

anonymous · Answer

@micahwood50 That does make sense. 
But it does not prove why $$\frac{ 1 }{ 2 }$$ does work in the equation. 
Like i said i think the statement X>1 is false with the equation given.

anonymous · Answer

i think we need to get like terms

anonymous · Answer

Never mind, I didn't realize the statement "for all x>1"...

anonymous · Answer

@Agentjamesbond007 
It seems as if you need to refine your statement "for all x>1" for us to solve it.

agentjamesbond007 · Answer

Well, that is the prompt I'm needed to prove. I'm not sure how to rephrase it.

anonymous · Answer

$$1/x = x ^{-1}$$
So using that you will have
$$2√x > 3 -x ^{-1}$$
You then will be able to square both sides and take the x's over.

anonymous · Answer

$$2\sqrt{x}>3-\frac{1}{x}$$We can assume $x$ is positive, so we will multiply both sides by $x$.$$2\sqrt{x^3}>3x-1$$We can then square both sides.  Since we know $x>1$, we know both sides are positive right now and greater than 1, so the equality will hold.$$4x^3>9x^2-6x+1$$$$4x^3-9x^2+6x-1>0$$Now we factor:$$(x-1)(x-1)(4x-1)>0$$As you can see, the only roots are $1$ and $\frac{1}{4}$.  These are the only times it crosses, the x-axis and therefore the only times it can change sign.  Since we know $x>1$, we know $x$ will never cross the x-axis.  So if it is positive for say $x=2$, then the inequality holds for all $x>1$.$$((2)-1)((2)-1)(4(2)-1)>0$$$$(1)(1)(7)>0$$$$7>0$$