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If either upper norm bound is nonpositive, there are no admissible functions with $\Vert f{\Vert }_q\ge C_q>0$, and the assertion is vacuous. For positive $C_p,C_r$, chooseSplitting the integral givesFor finite $r$, Holder inequality bounds the last term by $\Vert f{\Vert }_r^q\mu (A{)}^{1-q/r}\le C_r^q\mu (A{)}^{1-q/r}$. HenceFor $r=\infty$, the corresponding bound is ${\int }_A|f{|}^q\le C_r^q\mu (A)$. Thus an explicit choice proving the strict inequality isThis is a nonvanishing level-set bound between three Lp norms: the lower-$p$ bound prevents diffuse low-amplitude mass, and the higher-$r$ bound prevents arbitrarily concentrated spikes.

Work on $\mathbb{R}$ with Lebesgue measure. Without a uniform $L^p$ bound, useIt has $\Vert f_n{\Vert }_q=1$ and $\Vert f_n{\Vert }_r=n^{q/r-1}\le 1$, with $1/r=0$ for $r=\infty$. Each function still belongs to $L^p\cap L^r$, but its $L^p$ norm is unbounded. For every fixed $\epsilon >0$, its level set above $\epsilon$ is eventually empty, defeating any positive uniform $M$.

Without a uniform $L^r$ bound, use insteadNow $\Vert g_n{\Vert }_q=1$ and $\Vert g_n{\Vert }_p=n^{1-q/p}\le 1$, but for any fixed $\epsilon >0$ the high-amplitude level set has measure $n^{-q}\to 0$. These give the two distinct failures of the nonvanishing level-set bound between three Lp norms when either control is removed.