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Let $\alpha > 0$, use mathematical induction to prove that

$$\sqrt{\alpha+\sqrt{\alpha+\sqrt{\alpha+...+\sqrt{\alpha}}}} < \frac{1+\sqrt{4\alpha+1}}{2}$$

The square root sign appears n times on the left hand side.

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What did you try so far? – 5xum Sep 3 '14 at 10:54
i proved it for n = 1, assumed true for n = p, but couldnt prove it for n = p+1 – user157716 Sep 3 '14 at 10:55
up vote 3 down vote accepted

Let $b=\frac{1+\sqrt{4\alpha+1}}{2}$. A key feature for $b$ is that $b^2-\alpha=b$ or $b^2=\alpha+b$.

Now, let $a_n$ be the your LHS when there are $n$ square root signs. The base case $a_1<b$ is trivial. And the induction step follows because $$ a_{n+1}^2=\alpha+a_n<\alpha+b=b^2\implies a_{n+1}<b. $$

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Oh yes, If only i had seen the key feature in my exam :-( – user157716 Sep 3 '14 at 11:02
It was a hunch: at the induction step, you always have $a_{n+1}^2=\alpha+a_n<\alpha+b$, so I checked what relation there was between $\alpha+b$ and $b^2$. Good luck next time! – Kim Jong Un Sep 3 '14 at 11:04
yup... Thank you KimJongUn – user157716 Sep 3 '14 at 11:05
You could do a bit more: note $a_2>a_1$ and if $a_n>a_{n-1}$ then $a_{n+1}^2=\alpha+a_n>\alpha+a_{n-1}=a_n^2$ so that $a_{n+1}>a_n$. This means $\{a_n\}$ is an increasing sequence that is also bounded so a finite limit $x=\lim_na_n$ exists. This $x$ satisfies $x>0$ and $x^2=\alpha+x$, which together give $x=b$. Thus, $a_n$ can get arbitrarily close to $b$ if $n$ is large enough. – Kim Jong Un Sep 3 '14 at 11:22

HINT : For the $n=p+1$ step, all you need is to prove $$\frac{1+\sqrt{4\alpha +1}}{2}\gt \sqrt{\frac{1+\sqrt{4\alpha +1}}{2}}.$$

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but why? when n = p+1, i get p+1 square roots on the LHS, how can i use the assumption from there? – user157716 Sep 3 '14 at 11:00
Let $\beta=(1+\sqrt{4\alpha+1})/2$. We suppose that $\beta\gt f(p)$ where $f(p)$ is the left hand in the question. We have to prove that $\beta\gt f(p+1)$. But from the supposition, we have $\sqrt{\beta}\gt f(p+1)$. So, if we can prove that $\beta\gt \sqrt{\beta}$, we have $\beta\gt\sqrt{\beta}\gt f(p+1)$, so we can get $\beta\gt f(p+1)$, which is what we want. – mathlove Sep 3 '14 at 11:03

The first thing is to establish some notation so that you can apply induction.

Put $a_1=\sqrt{\alpha},a_n=\sqrt{\alpha+a_n}$. Check that that is the series you want.

Now you go about proving it by induction in the usual way. Start by showing that $a_1<(1+\sqrt{4\alpha+1})/2$. Then show that if $a_n<(1+\sqrt{4\alpha+1})/2$, so is $a_{n+1}$.

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