Prove that there exists no positive real number such that:

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.

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

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

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

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

To prove that there exists no positive real number that satisfies the inequality 4x^2 + 1/(4x) < 1, you need to use a different approach.

Let's start by assuming that there exists a positive real number, let's say 'a', that satisfies the given inequality:

4a^2 + 1/(4a) < 1

Rearrange the inequality to get:

4a^2 - 1 < -1/(4a)

Now, multiply both sides by 4a to eliminate the denominator:

16a^3 - 4a < -1

Rearrange the inequality again:

16a^3 - 4a + 1 < 0

Let's define a function f(a) = 16a^3 - 4a + 1. If we can prove that f(a) < 0 for all positive real numbers 'a', then we have proven that there exists no positive real number 'a' that satisfies the inequality.

To do this, we need to analyze the function f(a) in the interval (0, ∞). We can observe that f(a) is a polynomial function of degree 3, which means it is continuous.

Next, we can find the critical points of the function by taking the derivative:

f'(a) = 48a^2 - 4

Setting f'(a) equal to zero and solving for 'a', we get:

48a^2 - 4 = 0
a^2 = 4/48
a^2 = 1/12
a = ±√(1/12)

Since we are considering positive real numbers, we take the positive square root:

a = √(1/12)

By substituting this critical point into the original inequality:

f(√(1/12)) = 16(1/12)^(3/2) - 4(1/12) + 1

Evaluating this expression will give you a negative value, indicating that f(a) is negative for all positive real numbers 'a'.

Therefore, you can conclude that there exists no positive real number 'a' that satisfies the inequality 4x^2 + 1/(4x) < 1.