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I have to prove this inequality: $$ \forall n \in Z^+, \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2} $$

So far, I have done the base cases and assumed the inequality is true for some integer k, and I have gotten to the point where I need to show that: $$ \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} < \frac{\sqrt{k+1}}{2} $$ Can somebody point me in the right direction? I have tried squaring both sides, but because there is a + sign on the LHS, there are still square roots and it only makes things more complicated.

Is there a name for this inequality?

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  • $\begingroup$ Thanks for the help, but I am finding it hard to understand what you guys are talking about. If it is possible, please go step by step, forgive me if I am asking too much. Is there a name for this inequality? $\endgroup$
    – mib1413456
    Oct 24 '13 at 3:23
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I am not a fan of using (in)equality that you have to prove, since you have to be really careful what you do with that. So, let's try this:

\begin{align*} \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} &> \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+4} = \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2(k+2)} \\ &> \frac{\sqrt{k}}{2} + \frac{1}{2\sqrt{k+2}} = \frac{1}{2} \left(\sqrt{k} + \frac{1}{\sqrt{k+2}} \right) = \frac{1}{2} \sqrt{ k + 2\sqrt{\frac{k}{k+2}} + \frac{1}{k+2} } \\ &> \frac{1}{2} \sqrt{k + 1} \end{align*}

You'll need to justify why

$$2\sqrt{\frac{k}{k+2}} + \frac{1}{k+2} > 1,$$

But that should be really easy.

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I would look at your inequality this way: $$ \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}-\frac{\sqrt{k}}{2} . $$ The right hand side is $\frac1{2(\sqrt{k+1}+\sqrt{k})}$. So we have $$ \frac1{2(\sqrt{k+1}+\sqrt{k})}<\frac1{4\sqrt{k}}<\frac1{2\sqrt{k+1}}=\frac{\sqrt{k+1}}{2(k+1)}<\frac{\sqrt{k+2}}{2(k+1)}. $$

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  • $\begingroup$ I get your $$\frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}-\frac{\sqrt{k}}{2} .$$ part but I do not understand the part where you change it to $$\frac1{2(\sqrt{k+1}+\sqrt{k})}$$ $\endgroup$
    – mib1413456
    Oct 24 '13 at 1:19
  • $\begingroup$ @mib1413456 After combining the right into one fraction over $2$, multiply top and bottom by the conjugate $\sqrt{k+1}+\sqrt{k}.$ $\endgroup$
    – coffeemath
    Oct 24 '13 at 1:46
  • $\begingroup$ May I know how did you arrive at $$\frac1{2(\sqrt{k+1}+\sqrt{k})}<\frac1{4\sqrt{k}}<\frac1{2\sqrt{k+1}}=\frac{\sqrt{k+1}}{2(k+1)}<\frac{\sqrt{k+2}}{2(k+1)}.$$ So far, I only got $$\frac{\sqrt{k+2}}{2k+2} > \frac1{2(\sqrt{k+1}+\sqrt{k})}$$ $\endgroup$
    – mib1413456
    Oct 24 '13 at 2:44
  • $\begingroup$ This last inequality is the one you want to prove. What I did (to prove that inequality) is, in order of the inequality signs: 1) first $\sqrt{k+1}>\sqrt{k}$; 2) $2\sqrt{k}\geq\sqrt{k+1}$ (this is equivalent to $\sqrt{1+1/k}\leq2$); 3) $1/\sqrt{k+1}=(k+1)/\sqrt{k+1}$; 4) $\sqrt{k+1}<\sqrt{k+2}$. $\endgroup$ Oct 24 '13 at 2:52
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$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

\begin{align} {\large\sum_{i = 1}^{n}{\root{\vphantom{\large A}i + 1} \over 2i}} &= {1 \over 2}\sum_{i = 1}^{n}\pars{{1 \over \root{\vphantom{\large A}i + 1}} + {1 \over i\root{\vphantom{\large A}i + 1}}} > {1 \over 2}\sum_{i = 1}^{n}\pars{{1 \over \root{\vphantom{\large A}i + 1}} + {1 \over n\root{\vphantom{\large A}i + 1}}} \\[3mm]&= {1 \over 2}\,{n + 1 \over n}\sum_{i = 1}^{n} {1 \over \root{\vphantom{\large A}i + 1}} > {1 \over 2}\,{n + 1 \over n}\pars{n\,{1 \over \root{\vphantom{\large A}n + 1}}} \\[3mm]&={\large% {\root{n + 1} \over 2} > {\root{n} \over 2}} \end{align}

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