Inequality with Nested Radicals II

Problem

Inequality  with Nested Radicals II

Proof 1

Let $r_n=\sqrt{n+\sqrt{(n-1)+\sqrt{\ldots+\sqrt{2+\sqrt{1}}}}}.\;$ Obviously, $r_{n+1}=\sqrt{n+1+r_n},\;$ $r_1=1.$

We proceed by induction. For $n=2,\;$ the statement at hand takes place with equality:

$\sqrt{3}\le\sqrt{2+\sqrt{1}}\le\sqrt{3}.$

Suppose $\sqrt{n+1}\le r_n\le\sqrt{2n-1}\;$ and let's prove that

$\sqrt{n+2}\le r_{n+1}\le\sqrt{2n+1}.$

For the left inequality, observe that $r_{n+1}\ge\sqrt{n+2}\;$ is equivalent to $\sqrt{n+1+r_n}\ge\sqrt{n+2}\;$ which, in turn, is equivalent to $n+1+r_n\ge n+2,\;$ i.e., $r_n\ge 1,\;$ which is obviously true for $n\ge 2.$

For the right inequality, we have by the induction hypothesis that $r_{n+1}=\sqrt{r_n+n+1}\le\sqrt{\sqrt{2n-1}+n+1}.\;$ Suffice it to prove that $\sqrt{\sqrt{2n-1}+n+1}\le\sqrt{2n+1},\;$ or, after squaring, $\sqrt{2n-1}+n+1\le 2n+1,\;$ or equivalently, $\sqrt{2n-1}\le n.\;$ This reduces to the obvious $(n-1)^2\ge 0.$

Proof 2

For the right inequality, observe that, since $(a-1)^2\ge 0,\;$ $a^2\ge a+(a-1).\;$ Thus we successively have

$\displaystyle\begin{align} \sqrt{2n-1} &= \sqrt{n+(n-1)}\\ &=\sqrt{n+\sqrt{(n-1)^2}}\\ &\ge\sqrt{n+\sqrt{(n-1)+(n-2)}}\\ &=\sqrt{n+\sqrt{(n-1)+\sqrt{(n-2)^2}}}\\ &\ge\sqrt{n+\sqrt{(n-1)+\sqrt{(n-2)+(n-3)}}}\\ &\text{and so on.} \end{align}$

For the left inequality, $\sqrt{2+\sqrt{1}}\gt 1\;$ so that, for $n\ge 2,\;$ $\sqrt{(n-1)+\sqrt{\ldots+\sqrt{2+\sqrt{1}}}}\gt 1.\;$ Adding $n\;$ to both sides and taking the square root proves the left inequality.

Note that, for $n\gt 2,\;$ the inequalities are strict.

Proof 3

We know that the expression with the infinite number of radicals $\displaystyle\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}\;$ has a value $\displaystyle\frac{1+\sqrt{1+4n}}{2}.\;$ Since rather obviously $\sqrt{n+\sqrt{(n-1)+\sqrt{\ldots+\sqrt{2+\sqrt{1}}}}}\le\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}},\;$ all we need to prove the right inequality is to establish $\displaystyle\frac{1+\sqrt{1+4n}}{2}\le\sqrt{2n+1}.$

Squaring leads to an equivalent inequality: $2+4n+2\sqrt{1+4n}\le 8n+4,\;$ i.e., $\sqrt{1+4n}\le 2n+1,\;$ which is seen true by another round of squaring.

For the left inequality, observe that, for $n\ge 2,\;$ $\sqrt{(n-1)+\sqrt{\ldots+\sqrt{2+\sqrt{1}}}}\ge 1\;$ simply because already $\sqrt{n-1}\ge 1.\;$ Adding $n\;$ to both sides and taking the square root proves the left inequality.

Acknowledgment

Dorin Marghidanu has posted the problem and a solution (Proof 1) at the CutTheKnotMath facebook page and later added a solution (Proof 2) by Tran Van Hien.

 

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