Let $w$ denote the weight on $A$ so that $1-w$ is the weight on $B$. Recall from the properties of variance that

$\sigma_p^2 = w^2\sigma_A^2 + 2w(1-w)\sigma_A\sigma_B \rho_{AB}+ (1-w)^2\sigma_B^2$

Without loss of generality, assume $\sigma_A> \sigma_B$. We wish to show that

$w^2\sigma_A^2 + 2w(1-w)\sigma_A\sigma_B \rho_{AB}+ (1-w)^2\sigma_B^2\leq \sigma_A^2$

Note that

$\sigma_A^2 = \sigma_A^2 (w + (1-w)) ^2 = \sigma_A^2 w^2 + 2w(1-w)\sigma_A^2 + \sigma_A^2(1-w)^2$

Since $\sigma_A > \sigma_B$ and $w$, $(1-w)$, and $\sigma_A$ are positive, this means that

$\sigma_A^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B + \sigma_B^2(1-w)^2$

And since the correlation has the property that $-1 \leq \rho_{AB} \leq 1$ and $w$, $(1-w)$, $\sigma_B$ and $\sigma_A$ are all positive, it must be the case that

$\sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B + \sigma_B^2(1-w)^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B\rho_{AB} + \sigma_B^2(1-w)^2$

Therefore

$\sigma_A^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B\rho_{AB} + \sigma_B^2(1-w)^2$ $\square$

In words, looking at the formula for variance of a convex combination of random variables, the variance is maximized if the correlation between the assets is 1. In this case, the possible portfolio values as a function of $w$ are a straight line segment between $A$ and $B$, which clearly can't have a variance higher than either. Now, if the correlation is less than 1, then any combination of the two will be lower than the straight line case.

Intuitively, the returns to assets $A$ and $B$ will partially cancel each other out any time they are not a fixed multiple of each other. This canceling out behavior reduces the variance of the resulting portfolio. The worst-case scenario is that the two assets are equal to each other, so the portfolio can never have a higher variance than the component asst with the highest variance.

Let $w$ denote the weight on $A$ so that $1-w$ is the weight on $B$. Recall from the properties of variance that

$\sigma_p^2 = w^2\sigma_A^2 + 2w(1-w)\sigma_A\sigma_B \rho_{AB}+ (1-w)^2\sigma_B^2$

Without loss of generality, assume $\sigma_A \geq \sigma_B$. We wish to show that

$w^2\sigma_A^2 + 2w(1-w)\sigma_A\sigma_B \rho_{AB}+ (1-w)^2\sigma_B^2\leq \sigma_A^2$

Note that

$\sigma_A^2 = \sigma_A^2 (w + (1-w)) ^2 = \sigma_A^2 w^2 + 2w(1-w)\sigma_A^2 + \sigma_A^2(1-w)^2$

Since $\sigma_A \geq \sigma_B$ and $w$, $(1-w)$, and $\sigma_A$ are positive, this means that

$\sigma_A^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B + \sigma_B^2(1-w)^2$

And since the correlation has the property that $-1 \leq \rho_{AB} \leq 1$ and $w$, $(1-w)$, $\sigma_B$ and $\sigma_A$ are all positive, it must be the case that

$\sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B + \sigma_B^2(1-w)^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B\rho_{AB} + \sigma_B^2(1-w)^2$

Therefore

$\sigma_A^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B\rho_{AB} + \sigma_B^2(1-w)^2$ $\square$

In words, looking at the formula for variance of a convex combination of random variables, the variance is maximized if the correlation between the assets is 1. In this case, the possible portfolio values as a function of $w$ are a straight line segment between $A$ and $B$, which clearly can't have a variance higher than either. Now, if the correlation is less than 1, then any combination of the two will be lower than the straight line case.

Intuitively, the returns to assets $A$ and $B$ will partially cancel each other out any time they are not a fixed multiple of each other. This canceling out behavior reduces the variance of the resulting portfolio. The worst-case scenario is that the two assets are equal to each other, so the portfolio can never have a higher variance than the component asst with the highest variance.