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I am stuck on a proof where I need to demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$.

The proof provides me with the information that $x^2\gt 2$ and $x>0$.

I've taken the following steps to simplify the algebra...to the point where I arrive at:

$\frac{(x^2+2)^2}{(2*x)^2} = \frac {x^4+4*x^2+4}{4*x^2}$

I simplified this further to produce: $1 + \frac {1}{4}*x^2 + \frac{1}{x^2}$

In looking at this simplification, I see that I need to demonstrate (remembering the conditions), $\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$

Now, $\frac{1}{4}*x^2 \gt \frac{1}{2}$ and $\frac{1}{x^2}\lt \frac{1}{2}$ but I cannot figure out how to conclude that the addition of the two outputs something greater than 1. I imagine that I somehow need to show that the first term $\frac{1}{4}*x^2$ is somehow bigger than $0.5$ than $\frac{1}{x^2}$ is less than $0.5$.

In attempt to address this, I performed the following operation:

$\frac{\frac{1}{4}*x^2}{\frac{1}{x^2}} = \frac{1}{4}*x^4$

I know that $x^4$ must be greater than $4$...therefore $\frac{\frac{1}{4}*x^2}{\frac{1}{x^2}} \gt 1$.

Have I therefore successfully demonstrated that the addition of the two fractions:

$\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$

and can conclude that:

$(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$ ?

Edit: While the answers below also provide the solution to my question (albeit using a different strategy from the very beginning), I have figured out how to solve my question by finishing the procedure that I pursued above

Specifically, revisiting $\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$, let's subtract $1$ from either side and then multiply through by $4x^2$ to produce:

$x^4-4x^2+4 \gt 0$ (multiplying by $4x^2$ will not change the direction of the inequality because $x\gt0$ )

We can factor this to $(x^2-2)^2$ and because $x \gt 0$, this number will always be greater than $0$.

Community
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S.C.
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2 Answers2

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Note that

$$\frac{\frac{1}{4}\cdot x^2}{\frac{1}{x^2}} \gt 1 \implies \frac{1}{4}\cdot x^2-\frac{1}{x^2}>0$$

and then

$$\frac{1}{4}\cdot x^2+\frac{1}{x^2}>\frac{2}{x^2}$$

For the solution we can proceed as follows

$$\left(x-\left(\frac{x^2-2}{2 x}\right)\right)^2 \gt 2$$

$$\left(x^2-2x\left(\frac{x^2-2}{2x}\right)+\frac{(x^2-2)^2}{4x^2}\right) \gt 2$$

$$\frac{(x^2-2)^2}{4x^2} \gt 0 $$

which is true for $x^2\neq 2$ and $x\neq 0$.

user
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  • hmmm, sorry, but I am a little confused regarding the insight you have provided. I assumed that I should use your newly produced fraction to rewrite one of the initial statements, but then I arrived at $1+2/(x^2)$ which, for $x^2$ > 2 makes this statement always less than 2. Have a done something wrong? – S.C. Oct 04 '19 at 20:13
  • @S.Cramer I think there is a typo here $\frac{(x^2\color{red}+2)^2}{(2*x)^2} $ in the fourth line. – user Oct 04 '19 at 20:17
  • You're saying that it should be a negative sign? – S.C. Oct 04 '19 at 20:22
  • @S.Cramer It seems that the given inequality is equivalent to $\frac{(x^2-2)^2}{4x^2} \gt 0$ with minus sign. – user Oct 04 '19 at 20:24
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It is equivalent to $$\frac{(x^2-2)^2}{4x^2}>0$$

Dr. Sonnhard Graubner
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