Prove that for $n\ge 2$ the following inequality holds: $$\frac{1}{n+1}\left(1+\frac{1}{3}+\ldots +\frac{1}{2n-1}\right) >\frac{1}{n}\left(\frac{1}{2}+\ldots+\frac{1}{2n}\right).$$
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Tags: inequalities, induction, algebra proposed, algebra
professordad
22.04.2014 00:28
We use induction. For $n=2$ we have $\frac{4}{9} > \frac{3}{8}$, true.
Now assume that $\frac{1}{n+1} \cdot \left(1 + \frac{1}{3} + \cdots + \frac{1}{2n-1}\right) > \frac{1}{n} \cdot \left(\frac{1}{2} + \cdots + \frac{1}{2n}\right)$. Then,
\begin{align*} 1 + \frac{1}{3} + \cdots + \frac{1}{2n-1} &> \frac{n}{2(2n+1)} \\ \frac{1}{(n+1)^2 \cdot (n+2)} \cdot \left(1 + \frac{1}{3} + \cdots + \frac{1}{2n-1}\right) &> \frac{(2n+1)(n+2)-(2n+2)(n+1)}{(n+1)(n+2)(2n+1)(2n+2)} \\ \left[\frac{1}{n+2} - \frac{n}{(n+1)^2}\right] \cdot \left(1 + \frac{1}{3} + \cdots + \frac{1}{2n-1}\right) &> \frac{1}{(n+1)(2n+2)} - \frac{1}{(n+2)(2n+1)} \\ \frac{1}{n+2} \cdot \left(1 + \frac{1}{3} + \cdots + \frac{1}{2n+1}\right) &> \frac{n}{(n+1)^2} \cdot \left(1 + \frac{1}{3} + \cdots + \frac{1}{2n-1}\right) + \frac{1}{2} \cdot \frac{1}{(n+1)^2} \end{align*} Utilizing our inductive hypothesis we have
\begin{align*}\frac{1}{n+2} \cdot \left(1 + \frac{1}{3} + \cdots + \frac{1}{2n+1}\right) &> \frac{n}{(n+1)^2} \cdot \left[\frac{n+1}{n} \cdot \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n}\right)\right] + \frac{1}{2} \cdot \frac{1}{(n+1)^2} \\ \frac{1}{n+2} \cdot \left(1 + \frac{1}{3} + \cdots + \frac{1}{2n+1}\right) &> \frac{1}{n+1} \cdot \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} + \frac{1}{2n+2}\right) \end{align*} So the inequality holds for $n+1$ if it holds for $n$ and our induction is complete.
BSJL
22.04.2014 06:29
$ \frac{1}{n+1}*(1+\frac{1}{3}+...+\frac{1}{2n-1}) > \frac{1}{n}*(\frac{1}{2}+...+\frac{1}{2n}) $ $ \Leftrightarrow n(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{2n-1}-\frac{1}{2n}) > \frac{1}{2}+...+\frac{1}{2n} $ $ \Leftrightarrow n(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}) > \frac{1}{2}+...+\frac{1}{2n}$ $ \Leftrightarrow \frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+...+\frac{1}{2} > \frac{1}{2}+...+\frac{1}{2n}$ And the last one is trivial!
Sardor
14.05.2014 04:49
It's very easy problem and from Canada MO 1998.
sqing
14.05.2014 08:36
Sardor wrote: It's very easy problem and from Canada MO 1998. Canada 1998： Let $ n$ be a natural number such that $ n \geq 2$. Show that \[ \frac {1}{n + 1} \left( 1 + \frac {1}{3} + \cdot \cdot \cdot + \frac {1}{2n - 1} \right) > \frac {1}{n} \left( \frac {1}{2} + \frac {1}{4} + \cdot \cdot \cdot + \frac {1}{2n} \right). \] India Regional Mathematical Olympiad 1998
bobthesmartypants
17.03.2016 06:38
Let $H_n=\sum_{i=1}^n \dfrac{1}{i}$. Then we need to prove $$\dfrac{H_{2n}-\frac{1}{2}H_n}{n+1} > \dfrac{\frac{1}{2}H_n}{n}\implies \dfrac{H_{2n}-\frac{1}{2}H_n}{\frac{1}{2}H_n} > \dfrac{n+1}{n}\implies \dfrac{H_{2n}}{\frac{1}{2}H_n} > 2+\dfrac{1}{n}\implies \dfrac{H_n-H_n+H_{2n}}{\frac{1}{2}H_n} > 2+\dfrac{1}{n}\implies \dfrac{H_{2n}-H_n}{\frac{1}{2}H_n} > \dfrac{1}{n}$$$$\implies \dfrac{2n}{n+1}+\dfrac{2n}{n+2}+\cdots +\dfrac{2n}{2n} > \dfrac{1}{1}+\dfrac{1}{2}+\cdots +\dfrac{1}{n}$$which is obviously true.
Evenprime123
31.01.2018 19:06
Let \( S =\left(\frac{1}{2}+\ldots +\frac{1}{2n}\right)\). It suffices to prove: \( (\frac{1}{n+1})(S+\frac{1}{2}) > \frac{1}{n}S \Leftrightarrow (\frac{1}{n}-\frac{1}{n+1})S < \frac{1}{2n+2} \Leftrightarrow \frac{S}{n^2+n} < \frac{1}{2n+2} \Leftrightarrow S < \frac{n}{2} \) which is true because by GP sum formulas, we have \(S = 1-(\frac{1}{2})^n < 1 \leq \frac{n}{2}\), for \( n\geq 2\).