This is an attempt to strengthen your claim.

If $x$ is large then $B(x,y)\sim \Gamma(y)x^{-y}$ and hence $$B(\alpha x,\alpha)\sim \Gamma(\alpha)(\alpha x)^{-\alpha};$$ where $\Gamma(z)$ is the Euler Gamma function.

On the other hand, for small $\alpha$, we have the expansion $$\Gamma(1+\alpha)=1+\alpha\Gamma'(1)+\mathcal{O}(\alpha^2).$$ Since $\alpha\Gamma(\alpha)=\Gamma(1+\alpha)$, it follows that $$\Gamma(\alpha)\sim \frac1{\alpha}-\gamma+\mathcal{O}(\alpha)$$ where $\gamma$ is the Euler constant.

We may now combine the above two estimates to obtain $$\alpha x^{\alpha}B(\alpha x,\alpha)\sim \alpha x^{\alpha}\left(\frac1{\alpha}-\gamma\right)(\alpha x)^{-\alpha}=\left(\frac1{\alpha}-\gamma\right)\alpha^{1-\alpha}\geq1$$ provided $\alpha$ is small enough.

This is an attempt to strengthen your claim.

If $x$ is large then $B(x,y)\sim \Gamma(y)x^{-y}$ and hence $$B(\alpha x,\alpha)\sim \Gamma(\alpha)(\alpha x)^{-\alpha};$$ where $\Gamma(z)$ is the Euler Gamma function.

On the other hand, for small $\alpha$, we have the expansion $$\Gamma(1+\alpha)=1+\alpha\Gamma'(1)+\mathcal{O}(\alpha^2).$$ Since $\alpha\Gamma(\alpha)=\Gamma(1+\alpha)$, it follows that $$\Gamma(\alpha)\sim \frac1{\alpha}-\gamma+\mathcal{O}(\alpha)$$ where $\gamma$ is the Euler constant.

We may now combine the above two estimates to obtain $$\alpha x^{\alpha}B(\alpha x,\alpha)\sim \alpha x^{\alpha}\left(\frac1{\alpha}-\gamma\right)(\alpha x)^{-\alpha}=\left(\frac1{\alpha}-\gamma\right)\alpha^{1-\alpha}\geq1$$ provided $\alpha$ is small enough. For example, $0<\alpha<\frac12$ works.

This is an attempt to strengthen your claim.

If $x$ is large and then $B(x,y)\sim \Gamma(y)x^{-y}$ and hence $$B(\alpha x,\alpha)\sim \Gamma(\alpha)(\alpha x)^{-\alpha};$$ where $\Gamma(z)$ is the Euler Gamma function.

On the other hand, for small $\alpha$, we have $$\Gamma(\alpha)\sim \frac1{\alpha}-\gamma+\mathcal{O}(\alpha)$$ where $\gamma$ is the Euler constant.

We may now combine the above two estimates to obtain $$\alpha x^{\alpha}B(\alpha x,\alpha)\sim \alpha x^{\alpha}\left(\frac1{\alpha}-\gamma\right)(\alpha x)^{-\alpha}=\left(\frac1{\alpha}-\gamma\right)\alpha^{1-\alpha}\geq1$$ provided $\alpha$ is small enough.

This is an attempt to strengthen your claim.

If $x$ is large then $B(x,y)\sim \Gamma(y)x^{-y}$ and hence $$B(\alpha x,\alpha)\sim \Gamma(\alpha)(\alpha x)^{-\alpha};$$ where $\Gamma(z)$ is the Euler Gamma function.

On the other hand, for small $\alpha$, we have the expansion $$\Gamma(1+\alpha)=1+\alpha\Gamma'(1)+\mathcal{O}(\alpha^2).$$ Since $\alpha\Gamma(\alpha)=\Gamma(1+\alpha)$, it follows that $$\Gamma(\alpha)\sim \frac1{\alpha}-\gamma+\mathcal{O}(\alpha)$$ where $\gamma$ is the Euler constant.

We may now combine the above two estimates to obtain $$\alpha x^{\alpha}B(\alpha x,\alpha)\sim \alpha x^{\alpha}\left(\frac1{\alpha}-\gamma\right)(\alpha x)^{-\alpha}=\left(\frac1{\alpha}-\gamma\right)\alpha^{1-\alpha}\geq1$$ provided $\alpha$ is small enough.