First, you got the scaling wrong. The correct scaling for

$$ |x|^\alpha u(x) \lesssim \|u\|_{L^2}^{1-\beta} \|\nabla u\|_{L^2}^\beta $$

would be $\alpha = \frac{N}{2} - \beta$ where $N$ is the spatial dimension. So for smaller $\alpha$ you need more $\beta$, not less.

For $\beta \in [\frac12, 1]$ ($\beta = 1$ only works in $N > 2$) the desired inequality can be proven using essentially the same argument as what you gave for Strauss's Lemma.

Set $$ p := \frac{\beta + \frac{N}2 - 1}{\frac{N}2 - \beta} $$ Note that when $\beta = \frac12$ you have $p = 1$. And when $\beta = 1$ you have $p = \frac{N}{N - 2}$. And within this region you have $p \geq 1$.

and run the argument using instead of $ r^{N-1} u^2$, the function $ r^{N-1} |u|^{p+1} $ instead. The same argument you gave shows that

$$ r^{N-1} |u|^{p+1} \lesssim \int r^{N-1} |u|^{p} |\partial_r u| $$

Cauchy-Schwarz the RHS you get

$$ \lesssim \| u\|_{L^{2p}(\mathbb{R}^N)}^p \|\nabla u\|_{L^2} $$

This gives

$$ r^{\frac{N-1}{p+1}} |u| \lesssim \|u\|_{L^{2p}}^{\frac{p}{p+1}} \|\nabla u\|_{L^2}^{\frac{1}{p+1}} $$

The first term has $2p \in [2,\frac{2N}{N-2}]$ so by Gagliardo-Nirenberg-Sobolev inequality, can be bounded by

$$ \|u\|_{L^{2p}} \lesssim \|u\|_{L^2}^{\theta} \|\nabla u\|_{L^2}^{1-\theta} $$ for some $\theta \in [0,1]$. If you plug in the formula for $p$ in terms of $\beta$ above, you will find, after some routine algebra, the expression listed at the beginning of this answer.

First, you got the scaling wrong. The correct scaling for

$$ |x|^\alpha u(x) \lesssim \|u\|_{L^2}^{1-\beta} \|\nabla u\|_{L^2}^\beta $$

would be $\alpha = \frac{N}{2} - \beta$ where $N$ is the spatial dimension. So for smaller $\alpha$ you need more $\beta$, not less.

For $\beta \in [\frac12, 1]$ ($\beta = 1$ only works in $N > 2$) the desired inequality can be proven using essentially the same argument as what you gave for Strauss's Lemma.

Set $$ p := \frac{\beta + \frac{N}2 - 1}{\frac{N}2 - \beta} $$ Note that when $\beta = \frac12$ you have $p = 1$. And when $\beta = 1$ you have $p = \frac{N}{N - 2}$. And within this region you have $p \geq 1$.

and run the argument using instead of $ r^{N-1} u^2$, the function $ r^{N-1} |u|^{p+1} $ instead. The same argument you gave shows that

$$ r^{N-1} |u|^{p+1} \lesssim \int r^{N-1} |u|^{p} |\partial_r u| $$

Cauchy-Schwarz the RHS you get

$$ \lesssim \| u\|_{L^{2p}(\mathbb{R}^N)}^p \|\nabla u\|_{L^2} $$

This gives

$$ r^{\frac{N-1}{p+1}} |u| \lesssim \|u\|_{L^{2p}}^{\frac{p}{p+1}} \|\nabla u\|_{L^2}^{\frac{1}{p+1}} $$

The first term has $2p \in [2,\frac{2N}{N-2}]$ so by Gagliardo-Nirenberg-Sobolev inequality, can be bounded by

$$ \|u\|_{L^{2p}} \lesssim \|u\|_{L^2}^{\theta} \|\nabla u\|_{L^2}^{1-\theta} $$ for some $\theta \in [0,1]$. If you plug in the formula for $p$ in terms of $\beta$ above, you will find, after some routine algebra, the expression listed at the beginning of this answer.

On the other hand, there cannot be any estimate with $\beta < \frac12$. This can be seen by the following counterexample.

Let $u$ be a pulse around $r = 1$, with thickness $\epsilon$ and height 1.

$$ \|u\|_{L^2} \approx \epsilon^{1/2} $$

$$ \|\nabla u\|_{L^2} \approx \epsilon^{- 1/2} $$

Thus

$$ \|u\|_{L^2}^{1-\beta} \|\nabla u\|_{L^2}^\beta \approx \epsilon^{\frac12 - \beta} $$

If you take $\beta < \frac12$ and $\epsilon \searrow 0$, you get a sequence of functions with $\|u\|_{L^2}^{1-\beta} \|\nabla u\|_{L^2}^\beta \searrow 0$ but unit height, contradicting any possible control of $L^\infty$.