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.

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