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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!

 Nov 1, 2018
 #1
avatar+98182 
+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)

 

 

 

cool cool cool

 Nov 1, 2018
 #2
avatar+99352 
+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

 

-----------------------------------------------------------------------

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

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

 

 

------------------------------------------------------------

 

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

 Nov 3, 2018
edited by Melody  Nov 3, 2018
 #3
avatar+99352 
+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
 #4
avatar+98182 
+2

Good job, Melody  !!!!!

 

I tried to work on  both sides at once, but got nowhere.....your "one side at a time " approach is way  better..... [but definitely not brief....LOL!!!! ]

 

 

cool cool cool

CPhill  Nov 3, 2018
 #5
avatar+99352 
+1

Thanks Chris :)

Melody  Nov 4, 2018
 #6
avatar+39 
+1

Thanks guys!!!

 Nov 7, 2018

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