Greatest number under a constraint

Question

What is the largest number $M>0$ such that there are reals $k>0$ such that: $$\frac{M}3<M^{\frac{K-1}{K}}<\frac{2^{M^{\frac{1}{K}}}}{(\log_e2)M^{\frac{1}{K}}}?$$

Answer 1

There is no largest number $M>0$ with the required property, because the inequalities are strict. In fact, we have:

$$A=\{M>0: \exists k>0 : {M\over 3}< M^{k-1\over k} < {2^{M^{1\over k}}\over(\ln 2)M^{1\over k}}\}=\{M: 0<M<{8\over \ln2}\}$$

To prove this, let's first show that the open interval $(0,{8\over\ln2})$ is contained in $A$. So assume $0<M<{8\over\ln2}$, and let $\delta=8-(\ln2)M>0$. Choose a sequence of positive real numbers $k_n$ such that $$M^{1\over k_n}<3\,\forall n\enspace\hbox{and}\enspace\lim_{n\to\infty}M^{1\over k_n}=3\hskip 0.8cm (1)$$ Then, by continuity, $\lim_{n\to\infty}2^{M^{1\over k_n}}=2^3=8$, hence there exists a sufficiently large $n$ such that $$2^{M^{1\over k_n}}>8-\delta=(\ln2)M\hskip 2.1cm (2)$$ So for each $n$ for which (2) holds, we get by rearranging the inequalities (1) and (2): $${M\over 3}< M^{k_n-1\over k_n} < {2^{M^{1\over k_n}}\over(\ln 2)M^{1\over k_n}}$$ which means that $M\in A$. As for the other direction, assume now that $M\in A$. Then there exists some real $k>0$ such that ${M\over 3}<M^{1-{1\over k}}$, so $M^{1\over k}<3$, and also $(\ln 2) M<2^{M^{1\over k}}$. Since $2^x$ increases in $(0,\infty)$, we deduce: $(\ln 2)M<2^3=8$, so $M<{8\over\ln 2}$.

So our set is precisely the open interval $(0,{8\over\ln 2})$, and there is no "largest number" - i.e. a maximum -- but only a supremum, namely, ${8\over \ln2}$.