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!
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)
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
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.