Prove that the incircle is the smallest circle which passes through the three sides of a triangle

Consider areas.

Let $A,B,C$ be the vertices of the triangle with opposite sides measuring $a,b,c$. Pick any point $P$ within the triangle and render its distance to the $a$ side as $\alpha$ distance to the $b$ side as $\beta$ and distance to the $c$ side as $\gamma$. Draw the line segments connecting $P$ to the vertices thus dividing the area of the triangle into three smaller areas.

A circle centered on $P$ and touching all sides of the triangle must have a radius $r\ge\max(\alpha,\beta,\gamma)$. Meanwhile the area of $\triangle ABC$, being the sum of the smaller areas, is given by

$\color{blue}{S=(1/2)(a\alpha+b\beta+c\gamma)\le(1/2)(a+b+c)\max(\alpha,\beta,\gamma)\le(1/2)(a+b+c)r}$

Compare this with the specific result if $P$ is chosen as the incenter $I$, whereupon the distance to all three sides is the incircle radius $\rho$ and then the sum of the small triangle areas is

$\color{brown}{S=(1/2)(a\rho+b\rho+c\rho)=(1/2)(a+b+c)\rho}$

Comparing the brown equality with the blue inequality gives $r\ge\rho$, qed.