Prove that $\sqrt{6}-\sqrt{2}$ $> 1$.

Question

I'm trying to prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. However, I came up with this:

Since $\sqrt{4}=2$ and $\sqrt{9}=3$ we have a following inequality: $2 < \sqrt{6} < 3$.

On the other hand we have $\sqrt{1}=1$ and $\sqrt{4}=2$ so we have $1 < \sqrt{2} < 2$.

Let $n = \sqrt{6}-\sqrt{2}$,

Therefore $ 1<n<3$, which implies that $n>1$.

Is this a correct proof?

Thank you.

Answer 1

Unfortunately, no, your proof is not correct. The bound $\sqrt{6} > 2$ combined with the bound $\sqrt{2} < 2$ is too weak - consider $2.1 > 2$ and $1.9 < 2$, while their difference is far less than $1$. In fact, the best this can imply is that $\sqrt{6} - \sqrt{2} > 0$.

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For a different approach, notice that your inequality is equivalent to

$$\sqrt{6} > 1 + \sqrt{2}$$

Square both sides and see what happens.

Answer 2

We have $\sqrt 6-\sqrt 2>1$ if and only if $\sqrt 6>1+\sqrt 2.$ The square on right-hand side of the inequality is $1+2+2\sqrt 2=3+2\sqrt 2$. The remaining question is now whether $3\geq 2\sqrt 2$. Again by squaring this is equivalent to $9\geq 8$ which is valid.

Answer 3

How about something like this? Why don't we find out what needs to be true for the inequality to hold? If we end up some place we know is true, then all we need to do is start at the bottom and work backwards. So let's assume it's true and see what happens. $$\sqrt{6} - \sqrt{2} > 1$$ $$\Leftrightarrow \left(\sqrt{6} - \sqrt{2} \right)^2 > 1$$ $$\Leftrightarrow 6 - 2\sqrt{12} + 2 > 1$$ $$\Leftrightarrow 7 - 4\sqrt{3} > 0$$ $$\Leftrightarrow 7 > 4\sqrt{3}$$ $$\Leftrightarrow \frac{7}{4} > \sqrt{3}$$

Is this last line true? Well, of course. $\frac{7}{4} = 1.75$ and $\sqrt{3} = 1.73\cdots < 1.75$. Now, just start at the bottom (which we know is a true fact) and work your way up to end up at the inequality the question asks you to prove.

Answer 4

$\sqrt{6}-\sqrt{2}$ is positive because $\sqrt{x}$ is minimal at $0$ and always increasing at $x>0$.

Because $x^2$ is $1$ at $x=1$ and always increasing at $x>0$, numbers greater than 1 remain greater than $1$ when squared and $8-\sqrt{48}$ results from squaring $\sqrt{6}-\sqrt{2}\,$.

Since $8-\sqrt{49}=1$ and $\sqrt{49}$ exceeds $\sqrt{48}$ given properties of $\sqrt{x}$, it then follows that $\sqrt{6}-\sqrt{2} > 1$.

Answer 5

$$\sqrt{6}-\sqrt{2} = \frac{6-2}{\sqrt{6}+\sqrt{2}}= \frac{2}{\frac{\sqrt{6}+\sqrt{2}}{2}}> \frac{2}{\sqrt{\frac{6+2}{2}}}=1$$

$\bf{Added:}$

$$\sqrt{6} - \sqrt{2} > \sqrt{6.25} - \sqrt{2.25} = 2.5 - 1.5 = 1$$ since the function $x\mapsto \sqrt{x}$ is concave.