I am trying to prove the following:
Lemma. If a metric measure space $(X,d,\mu)$ supports a $p$-Poincaré inequality for some $p \in [1;\infty)$, then it supports a $q$-Poincaré inequality for all $q \geq p$.
Here, the Inequality is defined as
Definition. Let $p \in [1;\infty)$. A metric measure space $(X,d,\mu)$ supports a $p$-Poincaré inequality, if every ball in $X$ has positive and finite measure ant if there exist constants $C > 0$ and $\lambda \geq 1$ such that $$ \frac{1}{\mu(B)} \int_B |u(x)-u_B| \mathrm{d}\mu(x) \leq C \operatorname{diam}(B) \left( \frac{1}{\mu(\lambda B)} \int_{\lambda B} \rho(x)^p \mathrm{d}\mu(x) \right)^{\frac{1}{p}} $$ for every open ball $B$ in $X$, for every function $u : X \to \mathbb{R}$ that is integrable on balls, and for every upper gradient $\rho$ of $u$ in $X$. Here, $u_B := \frac{1}{\mu(B)} \int_B u(x) \mathrm{d}\mu(x)$ denotes the mean of $u$ in $B$, and $\operatorname{diam}(B):=\sup\{d(x,y) : x,y \in B\}$ the diameter of $B$.
In order to prove this, it should suffice to show $$ \left( \frac{1}{\mu(\lambda B)} \int_{\lambda B} \rho(x)^p \mathrm{d}\mu \right)^{\frac{1}{p}} \leq C' \left( \frac{1}{\mu(\lambda B)} \int_{\lambda B} \rho(x)^q \mathrm{d}\mu \right)^{\frac{1}{q}} $$ for some $C' \geq 1$.
EDIT: Solution
Choosing $a=\frac{q}{p}$ and $b=\frac{q}{q-p}$ we have $\frac{1}{a}+\frac{1}{b}=1$ and applying Hölder's Inequality $\int_{\lambda B} fg \mathrm{d}\mu \leq \left(\int_{\lambda B} f^a \mathrm{d}\mu\right)^{\frac{1}{a}} \left(\int_{\lambda B} g^b \mathrm{d}\mu\right)^{\frac{1}{b}}$ to $f:=\rho^p$ and $g:=1$ (as functions on $\lambda B$) I obtain $$ \begin{align*} \left( \frac{1}{\mu(\lambda B)} \int_{\lambda B} \rho^p \mathrm{d}\mu \right)^{\frac{1}{p}} &\leq \left( \frac{1}{\mu(\lambda B)} \left( \int_{\lambda B} (\rho^p)^{a} \mathrm{d}\mu \right)^{\frac{1}{a}} \left( \int_{\lambda B} 1^{b} \mathrm{d}\mu \right)^{\frac{1}{b}} \right)^{\frac{1}{p}} \\ &= \frac{1}{\mu(\lambda B)^{\frac{1}{p}}} \left( \int_{\lambda B} \rho^q \mathrm{d}\mu \right)^{\frac{1}{q}} \left( \int_{\lambda B} 1^{b} \mathrm{d}\mu \right)^{\frac{1}{bp}} \\ &= \frac{\mu(\lambda B)^{\frac{1}{q}}}{\mu(\lambda B)^{\frac{1}{p}}} \left( \frac{1}{\mu(\lambda B)} \int_{\lambda B} \rho^q \mathrm{d}\mu \right)^{\frac{1}{q}} \left( \int_{\lambda B} 1 \mathrm{d}\mu \right)^{\frac{1}{bp}} \\ &= \mu(\lambda B)^{\frac{1}{q} - \frac{1}{p} + \frac{1}{bp}} \left( \frac{1}{\mu(\lambda B)} \int_{\lambda B} \rho^q \mathrm{d}\mu \right)^{\frac{1}{q}} \\ &= \mu(\lambda B)^{0} \left( \frac{1}{\mu(\lambda B)} \int_{\lambda B} \rho^q \mathrm{d}\mu \right)^{\frac{1}{q}} \\ &= \left( \frac{1}{\mu(\lambda B)} \int_{\lambda B} \rho^q \mathrm{d}\mu \right)^{\frac{1}{q}} \end{align*} $$
You missed the relation $$ \frac1q-\frac1p + \frac1{bp}=0 $$ which follows from $b:=\frac{q}{q-p}$.