Prove that there exists no positive real number such that:

Question

4x^2+1/4x <1

Can I simplify this inequality in the following way:

multiply both sides by 4x so we have.

4x^2+1 <4x+1

Which renders the original statement false as 4x^2+1 cant be less than 4x+1.

Even further can I divide both sides by 4x to leave the following:

x+1 < x+1

Which again proves the statement cant be true as its the same on either side.

Does this work? im thinking outside the box here.

Answer 1

Some strange math taking place here.

first of all, the way you typed it ...
4x^2+1/4x <1 , if you multiply by 4x you would get 16x^4 + 1 < 4x

According to your result, I suspect
that your question was:
(4x^2+1)/(4x) < 1

then 4x^2 + 1 < 4x+1
4x^2 - 4x + 1 < 0
(2x + 1)^2 < 0
which of course is not possible since the square of anything cannot be negative.

I have no idea what you did to get
x+1 < x+1

Answer 2

4x^2+1/4x <1

Can I simplify this inequality in the following way:

multiply both sides by 4x so we have.

4x^2+1 <4x+1
==========================
I have no idea what you have and what you are doing.
First of all, whatever, No
If you have
(4x^2+1)/4x <1 NOTE-parentheses VITAL
and multiply both sides by 4 x
you have
4 x^2 + 2 < 4 x
that means
4 x^2 -4x + 2 <0
graph the parabola, where is the vertex?
y = 2 x^2 -2x +1
x^2 - x = y-.5
x^2 - x + 1/4 = y -.25
(x-1/2)^2= y-.25
vertex at (1/2,1/4) ABOVE x axis
so no negative y values
the end

Answer 3

looks like both Damon and I have a typo

in mine: from 4x^2 - 4x + 1 < 0
(2x - 1)^2 < 0
the conclusion is still the same

in Damon's
from
(4x^2+1)/4x <1 NOTE-parentheses VITAL
and multiply both sides by 4 x

you would get:
4x^2 + 1 < 4x <---- which is the same as mine

Answer 4

Yes, however you do it, the vertex is above the x axis :)