Real number inequality

Question

Suppose $a,b,c\in\mathbb{R}_{\geq 0}$ and $p,q\in\mathbb{Q}\cap[0,1]$ are fixed with $p+q=1$. Is it necessarily true that $a\leq b^pc^q$ implies $(a+\varepsilon)\leq(b+\varepsilon)^p(c+\varepsilon)^q$ for any $\varepsilon>0$?

Answer 1

Let $r=\frac{1}{p}$ and $s=\frac{1}{q}$ (the case when either $p$ or $q$ is zero can be checked manually). Then, $\frac{1}{r}+\frac{1}{s}=1$ and so by Holder's inequality, we have: $$ (b+e)^p(c+e)^q=[(b^{1/r})^r+(e^{1/r})^r]^{1/r}[(c^{1/s})^s+(e^{1/s})^s]^{1/s}\\ \geq(b^{1/r}c^{1/s})+e^{1/r}e^{1/s}=b^pc^q+e^{p+q}\geq a+e. $$