Prove this inequality

Problem 1.

Prove that the inequality

k=1n(1+13k2)>3n+13\prod_{k=1}^n \left( 1+\frac{1}{3k-2}\right)>\displaystyle \sqrt[3]{3n+1}

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

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

Proof:

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

mM<m+cM+cMm>M+cm+c\begin{aligned}\frac{m}{M} &< \frac{m+c}{M+c} \\ \frac{M}{m} &> \frac{M+c}{m+c} \end{aligned}

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

By Lemma 1,

k=1n(1+13k2)=k=1n(3k13k2)>k=1n(3k3k1)>k=1n(3k+13k)\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,

(k=1n(3k13k2))2>k=1n(3k3k1)k=1n(3k+13k)=k=1n(3k+13k1)=(3n+1)k=1n1(3k+1)k=1n(3k1)1=(3n+1)k=1n(3(k1)+1)k=1n(3k1)1=(3n+1)k=1n(3k23k1)\begin{aligned}\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{aligned}

Hence

(k=1n(3k13k2))3>3n+1 \left(\prod_{k=1}^n \left(\frac{3k-1}{3k-2}\right)\right)^3 > 3n+1

since x3\sqrt[3]{x} is an increasing function

k=1n(1+13k2)>3n+13\prod_{k=1}^n \left( 1+\frac{1}{3k-2}\right) > \sqrt[3]{3n+1}

and this completes our proof.

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

Note by Daniel Xiang
1 year, 8 months ago

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