Diversification is key.
The clear cut answer is diversification. A weighted combination of assets will more often than not show a lower return variance than even the asset with the lowest variance across the asset universe.
The setup
Without loss of generality, let us assume there exist two assets $a$ and $b$ with variance $\sigma_a^2=\alpha^2<\sigma_b^2=1$. These assets are correlated with parameter $\rho \in [0,1]$.
From basic portfolio theory we know that asset weights in the minimum-variance-portfolio are
$$
w_{MVP}=\frac{\Sigma^{-1}\mathbf{1}}{\mathbf{1}^T\Sigma^{-1}\mathbf{1}}
$$
with $\mathbf{1}$ a vector of ones. In our setup, $ \Sigma = \begin{pmatrix}\alpha^2 & \alpha\rho \\ \alpha\rho & 1\end{pmatrix} $ and thus the optimal weight on asset $a$ is
$$
w_a\equiv w_{MVP,a}=\frac{1-\alpha\rho}{1+\alpha^2-2\alpha\rho}
$$
We can now answer some questions.
When is the weight on the asset with lower risk exactly equal to 100%?
The weight on asset $a$ is 100% if
$$
\begin{align}
1&\stackrel{!}{=}w_a\\
&=\frac{1-\alpha\rho}{1+\alpha^2-2\alpha\rho}\\
\Rightarrow 1+\alpha^2-2\alpha\rho &= 1-\alpha\rho\\
\Rightarrow \alpha&=\rho
\end{align}
$$
i.e. when asset $a$'s volatility in relationship to asset $b$'s volatility (conveniently, $\alpha$) equals its correlation with asset $b$.
When will total variance be smaller than the smallest asset variance, $\sigma_a^2=\alpha^2$?
The total variance of the MVP equals
$$
\sigma_{MVP}^2=\frac{1}{\mathbf{1}^T\Sigma^{-1}\mathbf{1}}=\frac{\alpha^2(1-\rho^2)}{1+\alpha^2-2\alpha\rho}
$$
It is smaller than the smallest asset variance $\sigma_a^2=\alpha^2$ if:
$$
\begin{align}
\alpha^2 &\stackrel{!}{>} \sigma_{MVP}^2\\
&=\frac{\alpha^2(1-\rho^2)}{1+\alpha^2-2\alpha\rho}\\
\Rightarrow 1+\alpha^2-2\alpha\rho &> 1-\rho^2\\
\Rightarrow \alpha^2-2\alpha\rho+\rho^2 &> 0 \\
\Rightarrow \left(\alpha-\rho\right)^2 &> 0
\end{align}
$$
The last expression is a quadratic form in $\alpha,\rho$. Hence, any combination of $\alpha,\rho$ with $\alpha \neq \rho$ will result in a decrease in total variance.