# Prove this inequality

Problem 1.

Prove that the inequality

$\prod_{k=1}^n \left( 1+\frac{1}{3k-2}\right)>\displaystyle \sqrt[3]{3n+1}$

is true $$\forall n \in \mathbb{Z}^{+}$$

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Proof:

Lemma 1: if the real numbers $$M, m, c>0$$ satisfy $$M > m$$, then

\begin{align}\frac{m}{M} &< \frac{m+c}{M+c} \\ \frac{M}{m} &> \frac{M+c}{m+c} \end{align}

The lemma is self-explanatory if we think of $$\frac{m}{M}$$ as the "saltiness" of a glass of salty water with total mass $$M$$, and $$m$$ is the mass of the salt. By increasing the numerator and denominator by the same amout (adding salt of mass $$c$$ in the glass) the "saltiness" increases.

By Lemma 1,

$\prod_{k=1}^n \left( 1+\frac{1}{3k-2}\right) = \prod_{k=1}^n \left(\frac{3k-1}{3k-2}\right) > \prod_{k=1}^n \left(\frac{3k}{3k-1}\right)>\prod_{k=1}^n \left(\frac{3k+1}{3k}\right)$

Therefore,

\begin{align}\left(\prod_{k=1}^n \left(\frac{3k-1}{3k-2}\right)\right)^2> \prod_{k=1}^n \left(\frac{3k}{3k-1}\right)\prod_{k=1}^n \left(\frac{3k+1}{3k}\right)&=\prod_{k=1}^n \left(\frac{3k+1}{3k-1}\right)\\&=(3n+1)\prod_{k=1}^{n-1}(3k+1) \prod_{k=1}^n \left( 3k-1\right)^{-1}\\&=(3n+1)\prod_{k=1}^{n}(3(k-1)+1) \prod_{k=1}^n \left( 3k-1\right)^{-1} \\& = (3n+1)\prod_{k=1}^n \left(\frac{3k-2}{3k-1}\right) \end{align}

Hence

$\left(\prod_{k=1}^n \left(\frac{3k-1}{3k-2}\right)\right)^3 > 3n+1$

since $$\sqrt[3]{x}$$ is an increasing function

$\prod_{k=1}^n \left( 1+\frac{1}{3k-2}\right) > \sqrt[3]{3n+1}$

and this completes our proof.

=============================================

Note by Daniel Xiang
9 months, 2 weeks ago

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