Prove that : $\dfrac{n+1}{2} \leq 2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}$.
I am unable to prove this even by induction and general method. Indeed, when I look at the question $2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}\leq n+1$, asked by me, I have received a hint as a comment to use binomial theorem and showed $$\left(1+\frac1n\right)^n=\sum_{k=0}^n{n\choose k}\frac1{n^k}\geq1+{n\choose 1}\frac1n=2.$$ So, expression becomes $$2 \cdot \sqrt{2} \cdot \sqrt[3]{2} \cdots \sqrt[n]{2} \leq \left(1 +\dfrac{1}{1}\right)\left(1 +\dfrac{1}{2}\right)\left(1 +\dfrac{1}{3}\right)...\left(1 +\dfrac{1}{n}\right)=n+1.$$ I want to solve this problem exactly by same method. So, for this I've to prove that $$\sqrt[n]{2} \leq \dfrac{n+2}{n+1}.$$ How do I do this ?
Considering the rhs, $$A_n=\prod_{i=1}^n 2^{\frac 1 n}$$ $$\log(A_n)=\sum_{i=1}^n \log(2^{\frac 1 n})=\log(2)\sum_{i=1}^n \frac 1 n=H_n\log(2)$$ For large values of $n$, $$H_n=\gamma +\log(n)+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ while the logarithm of the lhs would write $$\log(2)+\log(n) +\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ and, using these short expansions, you could see that $\left(\log(rhs)-\log(lhs)\right)$ cancels close to $n=32$ and from that point, becomes negative.
So, as comments already showed it, the inequality only holds for a small range of $n$ (up to some finite $n$ as Did properly pointed out).