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

Question

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.

Answer 1

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 $$ a_{n+1}^2=\alpha+a_n<\alpha+b=b^2\implies a_{n+1}<b. $$

Answer 2

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

Answer 3

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 $a_{n+1}$.