If$0 < q < p < 1$and $\lambda > 0$, there is $\delta(\lambda)$ s. t. $\frac{t^q}{(t + \varepsilon)^{q + \beta}} \geq \lambda t^p$ for $0 < t < \delta$ The following is a claim in a paper I am reading (Montenegro, M. - Existence of Solutions to a Singular Elliptic Equation - Milan Journal of Math. 2011):
If $0 < q < p < 1$ and $\lambda > 0$, there exists $\delta(\lambda)$ such that $\frac{t^q}{(t + \varepsilon)^{q + \beta}} \geq \lambda t^p$ for all $0 \leq t \leq \delta$, where $0 < \varepsilon, \beta < 1$.
I have some intuition on taking $\delta$ so small that the denominator of the fraction is dominated by $\varepsilon$, bu cannot formalize a proper proof.
Any hints will be the most appreciated.
Thanks in advance and kind regards.
 A: Let $ \ f: [0, \infty[ \, \to \mathbb{R} \ $ be such that $$f(t) = \lambda \, t^{p-q} (t + \varepsilon )^{q+ \beta} \ \ , $$ for all $ \ t \geq 0 \ $. It is straightforward that $f$ is continuous. So $f$ is continuous at $ \ 0 \in \mathbb{R} \ $. Then $ \ \lim_{t \to 0^+} f(t) = f(0) \ $. Since $ \ 1>0 \, $, there must exist $ \ \tilde{\delta} > 0 \ $ such that, for all $ \ t \in \mathbb{R} \, $, $$0 < t < \tilde{\delta} \ \Longrightarrow \ |f(t) - f(0)|<1 \ \ . $$
Take $ \ \delta = \tilde{\delta}/2 \ $ and notice that $f$ is nondecreasing and that $ \ f(0)=0 \ $ to get $$0 \leq t \leq \delta \ \Longrightarrow \ 0 \leq f(t) \leq 1 \ \ . $$
A: since $t\mapsto t^{p-q}(t+\epsilon)^{q+\beta}$ is strictly increasing, then the question is equivalent to the existence of $\delta=\delta(\lambda)$ such that
$$\frac{1}{\lambda}\geq \delta^{p-q}(\delta+\epsilon)^{q+\beta}.\qquad (1)$$
Such $\delta$ is the solution of the inequality
$$\frac{1}{\lambda}\geq \delta^{p-q}(\delta+\epsilon)^{q+\beta}>\delta^{p-q} \delta^{q+\beta}=\delta^{p+\beta}\qquad (2)$$
Hence
$$\delta<\frac{1}{\lambda^{\frac{1}{p+\beta}}}$$.
And since $\beta+p<2$, then $\delta^{p+\beta}\geq \delta^2$ provided $\delta\leq 1$. That is we have
$$\delta<min\{1,\frac{1}{\lambda^{\frac{1}{2}}}\}$$
