First, we homogenize your inequality. I.e., we plug in $\lambda u$ with $\lambda > 0$ and consider $\lambda \to 0$ and $\lambda \to \infty$. This shows that the inequality cannot hold in case $2p \not\in [r,2]$.
In case $2p = r$, the first term on the rhs disappeards and your inequality is equivalent to
$$ \|u\|_2 \lesssim \|u\|_{1,r}.$$
This holds under some relation of $r$ and the dimension $d$ (Sobolev embedding).
In the remaining case $2 p \in (r,2)$, you can divide by $\lambda^{2p}$ and minimize the resulting rhs w.r.t. $\lambda > 0$. This results in the homogeneous inequality
$$
\|u\|_2 \lesssim \|u\|_{1,2}^{2\frac{1-r/(2p)}{2-r}} \|u\|_{1,r}^{r\frac{2/(2p)-1}{2-r}}.
$$
Note that Young's inequality implies that this inequality is indeed equivalent to your original inequality;
and the exponents
on the right-hand side add up to one.
Now, we can try to produce this inequality
by combining Sobolev embedding with interpolation (via Hölder's inequality).
From Sobolev, we get
$$
\|u\|_q \lesssim \|u\|_{1,2},
\qquad
\|u\|_s \lesssim \|u\|_{1,r}
$$
for
$ 1/q = 1/2 - 1/d$
and
$ 1/s = 1/r - 1/d$.
Thus,
$$
\|u\|_{q}^{2\frac{1-r/(2p)}{2-r}} \|u\|_{s}^{r\frac{2/(2p)-1}{2-r}}
\lesssim
\|u\|_{1,2}^{2\frac{1-r/(2p)}{2-r}} \|u\|_{1,r}^{r\frac{2/(2p)-1}{2-r}}.
$$
Interpolation implies
$$
\|u\|_t
\lesssim
\|u\|_{1,2}^{2\frac{1-r/(2p)}{2-r}} \|u\|_{1,r}^{r\frac{2/(2p)-1}{2-r}}.
$$
for
$$
\frac1t
=
\frac{2\frac{1-r/(2p)}{2-r}}{s}
+
\frac{r\frac{2/(2p)-1}{2-r}}{q}
.
$$
Finally,
we have
$$
\|u\|_2
\le
\|u\|_t
$$
if $2 \le t$.
Hence, in case
$$
\frac12
\ge
\frac{2\frac{1-r/(2p)}{2-r}}{s}
+
\frac{r\frac{2/(2p)-1}{2-r}}{q}
.
$$
we have shown your inequality.
(If this relation is not satisfied, I expect that your inequality does not hold)
