An inequality involving n-th real root of real numbers

Question

Given any $1\leq K\leq a$ and $1\leq K\leq b^{1/n}, $ $n$ is any positive integer, I was trying to prove

$$\frac{1}{1+a}+\frac{1}{1+b^{1/n}}-\frac{1}{1+(ab)^{1/(n+1)}}\leq \frac{1}{1+K},$$ where $x^{1/n}$ represents nth real root of $x.$ Will the induction principle help?

Answer 1

The RHS of the inequality is decreasing with $K$, so for inequality to hold in general it is enough that it holds for the minimum value of the RHS, which is attained for the maximum $K=\sqrt[n]{b}\,$:

$$\require{cancel} \begin{align} \frac{1}{1+a}+\cancel{\frac{1}{1+\sqrt[n]{b}}}-\frac{1}{1+\sqrt[n+1]{ab}}\leq \cancel{\frac{1}{1+\sqrt[n]{b}}} \;\;&\iff\;\; \frac{1}{1+a} \le \frac{1}{1+\sqrt[n+1]{ab}} \\ &\iff\;\; \cancel{1}+\sqrt[n+1]{ab} \le \cancel{1} + a \\ &\iff\;\; \bcancel{a}b \le a^{n+\bcancel{1}} \\ &\iff\;\; \sqrt[n]{b} \le a \end{align} $$

Therefore the posted inequality does not hold in general, but the following reformulation does:

Given $1\leq K\leq \sqrt[n]{b} \le a\,$, $n \in \mathbb{N}\,$: $\;\;\displaystyle\frac{1}{1+a}+\frac{1}{1+\sqrt[n]{b}}-\frac{1}{1+\sqrt[n+1]{ab}}\leq \frac{1}{1+K}\,$.