Prove that \(\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}\) for 0 < x < 1.

Please provide a full solution.

Thanks for the help!

FencingKat Nov 1, 2018

#1**+3 **

Don't know if this is what you are looking for....but....here's a solution using Calculus

The first function can be written as

y = (1/√2) ( 2x^2 - 2x + 1)^(1/2)

The derivative of this function is

y' = ( 1/ sqrt(8)) ( 2x^2 - 2x +1)^(-1/2) ( 4x - 2)

We only need to solve this to find a possible x value of a max / min of this function [a critical point ]

4x - 2 = 0

4x = 2

x = .5 = 1/2

And the y value at this point = [1/ √2] ( 2(1/2)^2 - 2(1/2) + 1)^(1/2) =

[1/√2 ] ( 1/2 - 1 + 1 )^(1/2) = (1/4)^(1/2) = 1/2

This point is a min on [ 0, 1] because

And at x = 0 the function value = 1/√2 ≈ .707

And at x = 1 the function value = 1/√2 ≈ .707

So...the point (.5, .5) is the lowest point on this graph from (0, 1)

The second function can be written as

y = x / ( x^2 + 1) = x ( x^2 + 1)^(-1)

The derivative of this is

y ' = (x^2 + 1)^(-1) - x (x^2 + 1)^(-2) * 2x factor this

y ' = (x^2 + 1)^(-2) [ (x^2 + 1 - 2x^2 ]

y' = ( x^2 + 1)^(-2) [ 1 - x^2)

To find the possible x that minimizes / maximizes this function on the interval we can solve this

1 - x^2 = 0

(1 - x) ( 1 + x) = 0 set both factors to 0 and solve for x and we get that

x = 1 or x = -1 and these are the critical points of this function

We can ignore the second answer since it's out of the interval

So...the function has either a max or min value at x = 1

At x = 1 the y value is 1 / [ 1^2 + 1] = 1/2

At x = 0 the y value is 0 / [ 0^2 + 1] = 0

Since the value a x = 1 is greater than at x = 0....this fuction has a max value on [ 0, 1] of 1/2....and at all other points in (0,1) y is less than 1/2

But the first function has a * minimum* y value of 1/2 on (0,1)

So..on the interval (0,1) the first function is always greater than the second function

So

√[ (2x^2 - 2x + 1) / 2 ] ≥ 1 / [ x + 1/x ] on (0, 1)

CPhill Nov 1, 2018

#2**+3 **

Hi FencingKat,

I have been asked if I can show or prove this without using calculus. So here goes ....

Prove that

\(\sqrt{ \frac{2x^2 - 2x + 1}{2} } \ge \frac{1}{x + \frac{1}{x}}\qquad for \quad 0

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First I will consider the LHS, I am going to look at it bit by bit.

For the LHS to be real

\(2x^2-2x+1\ge0\)

consider

\(y=2x^2-2x+1 \\ \)

This is a concave up parabola so it will be negative between the roots.

First find the roots.

\(2x^2-2x+1 =0\\ x=\frac{2\pm\sqrt{4-8}}{4} \)

there are no solutions to this so the roots are imaginary.

This parabola is always above the x axis

\(2x^2-2x+1>0 \;\;for\;\;all\;\;real\;\;x \\ \)

The minimum value is at x=1/2 and it is

2*0.5^2-2*0.5+1 = 0.5

so for 0

\(\text{minimum value of }\sqrt{\frac{2x^2-2x+1}{2}}=\sqrt{\frac{0.5}{2}}=\frac{1}{2} \)

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Now consider the RHS

\(RHS = \frac{1}{x+\frac{1}{x}} \\\text{As x approaches } 0^+ \text{ the RHS approaches 0}\\ \text{As x approaches 1 the RHS approaches }\frac{1}{2}\)

Now I suspect that the RHS is never more than 0.5 in the given domain

so i am going to determine for what values of x is the RHS less then or equal to 0.5

\(\frac{1}{x+\frac{1}{x}}\le\frac{1}{2}\\ 2 \le x+\frac{1}{x}\\ 2x \le x^2+1\\ 0 \le x^2-2x+1\\ x^2-2x+1\ge0\\ \triangle=b^2-4ac=4-4=0\\ \text{There is only 1 root (which we already know is when x=1)}\\ \text{plus } y=x^2-2x+1 \text{ is a concave up parabola so }\\ \frac{1}{x+\frac{1}{x}}\le\frac{1}{2} \text{ for all positive real x} \)

So the maximum of the RHS is 0.5 and that occurs when x=1

and the minimum of LHS is 0.5 and that occurs when x= 0.5

so thereforeLHS is not just greater or equal to RHS for all x between 0 and 1

I can go further and say that

LHS is greater than RHS for all all real x greater than 0.

so it is proven that

\(\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}\qquad for \quad 0

it is also proven that

\(\sqrt{ \frac{2x^2 - 2x + 1}{2} } > \frac{1}{x + \frac{1}{x}}\qquad for \quad x>0\)

Here is a pic

Melody Nov 3, 2018

#3**+2 **

Sorry, I know this last post is not displaying quite properly.

I suspect that is becsause it has too many elements too it and that is causing problems.

But I think it is all readable.

Melody
Nov 3, 2018